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STRENGTH OF MATERIALS

A. COMPREHENSIVE PRESENTATION OF SCIENTIFIC METHODS

OF LOCATING AND DETERMINING STRESSES AND

CALCULATING THE REQUIRED STRENGTH

AND DIMENSIONS OF BUILDING

MATERIALS

EDWARD R. MAURER, B.C.E.

MECHANICS,

FBornssoB OF

UNIVERSITY OF WISCONSIN

ILLUSTRATED

AMERICAN TECHNICAL SOCIETY

CHICAGO

1917

COPYRIGHTED IN GREAT BRITAIN
ALL RIGHTS RESERVED

INTRODUCTION

EVERY layman is fascinated by a great engineering work — a
large sewage system, a Keokuk dam, a twenty-story steel
structure — and wishes he might be able to construct such work
and carry it to completion. And yet he hardly appreciates the
knowledge, experience, and judgment necessary to bring such a
work to a satisfactory close. Time was when all of the details of
such structures were determined by guesswork, but the develop-
ments in science and mathematics have changed all that. For-
mulas for the various types of stresses have been worked out;
constants for every known material have been collected; and a
multitude of diagrams and tables contribute to making the engi-
neer's work as precise as a bookkeeper's balance. The strength
and size of every rivet and the length and cross-section of every
girder in a steel structure are figured so they will bear the strains
put upon them. The design of a masonry dam, the strength of
the concrete mixture, and the amount of steel reinforcement are
all mathematically determined in order to safely restrain the given
volume of water behind it.

<I The final judgment, therefore, as to the size of every part of
the structure — must depend upon the designer's knowledge of
"Strength of Materials." The treatment of Strength of Materials,
as found in standard textbooks, is so clothed in abstruse mathe-
matics that it is impossible for the average trained man to obtain
a working knowledge of the subject. As the subject is one of the
most important in any of the engineering branches, the author
has thought it wise to present it in simple form, culling out all
material that can not be used in practical design and analyzing
the subject from the theory through to the practical formulas
without the use of higher mathematics. In fact, he has used only
such mathematics as may be easily understood. While the author
has designed this work especially for home study purposes, the
material is valuable to the college trained man as well, as it gives
in clear, concise form the principles which are most used in engi-
neering and architectural work. It is the hope of the publishers
that the book will fill a place among the useful reference works in
the field of engineering.

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CONTENTS

PAGE

Simple stress 1

Kinds of stress 2

Tension, compression, shear 2

Unit-stress 3

Deformation 4

Elasticity 4

Hooke's law, and elastic limit 5

Ultimate strength 6

Stress-deformation diagram 6

Working stress and strength 8

Factor of safety 8

Strength of materials in tension 11

Strength of materials in compression 13

Strength of materials in shear 15

Reaction of supports 16

Moment of a force 16

Principle of moments 17

Kinds of beams 19

Determination of reactions on beams 19

External shear and bending moment 23

External shear 23

Rule of signs 23

Units for shears 24

Shear diagrams 27

Maximum shear 31

Bending moment 32

Moment diagrams 36

Maximum bending moment '. . . . 40

Table of maximum shears, moments, etc 41

Center of gravity and moment of inertia 41

Center of gravity of an area 41

Principle of moments applied to areas 42

Center of gravity of built-up sections 44

Moment of inertia 46

Table of centers of gravity and moments of inertia 52

Strength of beams 52

Kinds of loads 52

Transverse, longitudinal, inclined forces 52

Neutral surface 54

Neutral line 54

Neutral axis 54

Stress at cross-section 55

Fibre stress. . . .57

CONTENTS

Strength of beams (Continued) PAGE

Value of the resisting moment 58

First beam formula 59

Applications of the first beam formula 59

Laws of strength of beams 71

Modulus of rupture 72

Resisting shear 73

Second beam formula 73

Horizontal shear 75

Design of timber beams .'76

Kinds of loads and beams 79

Flexure and tension 79

Flexure and compression 81

Combined flexural and direct stress 83

Strength of columns 85

End conditions 85

Classes of columns 86

Cross sections of columns 86

Radius of gyration 86

Kinds of column loads 88

Rankine's column formula 88

Graphical representation of column formulas 93

Combination column formulas 94

Straight-line and Euler formulas 94

Parabola-Euler formulas 98

Broken straight-line formula 101

Design of columns 102

Strength of shafts 105

Twisting moment 105

Torsional stress 106

Resisting moment 107

Formula for the strength of a shaft 107

Formula for the power which a shaft can transmit 109

Stiffness of rods, beams, and shafts 110

Coefficient of elasticity Ill

Temperature stresses 113

Deflection of beams 114

Twist of shafts 116

Non-elastic deformation * 117

Riveted joints 118

Kinds of joints 118

Shearing strength, or shearing value, of a rivet 118

Bearing strength, or bearing value, of a plate 119

Frictional strength of a joint .- 119

Tensile and compressive strength of riveted plates 120

Computation of the strength of a joint 120

Efficiency of a joint 122

STRENGTH OF MATERIALS.

PART I.

SIMPLE STRESS.

i. Stress. When forces are applied to a body they tend in a
greater or less degree to break it. Preventing or tending to pre-
vent the rupture, there arise, generally, forces between every two
adjacent parts of the body. Thus, when a
load is suspended by means of an iron rod,
the rod is subjected to a downward pull at
its lower end and to an upward pull at its
upper end, and these two forces tend to pull
it apart. At any cross-section of the rod
the iron on either side "holds fast" to that on
the other, and these forces which the parts
of the rod exert upon each other prevent
the tearing of the rod. For example, in Fig.
1, let a represent the rod and its suspended
load, 1,000 pounds; then the pull on the
lower end equals 1,000 pounds. If we neg-
lect the weight of the rod, the pull on the

o s.

upper end is also 1,000 pounds, as shown in

Fig. 1 (5) ; and the upper part A exerts

on the lower part B an upward pull Q equal

to 1,000 pounds, while the lower part exerts

on the upper a force P also equal to 1,000 pounds. These two

forces, P and Q, prevent rupture of the rod at the "section" C; at

any other section there are two forces like P and Q preventing

rupture at that section.

By stress at a section of a body is meant the force which the
part of the body on either side of the section exerts on the other.
Thus, the stress at the section C (Fig. 1) is P (or Q), and it equals
1,000 pounds.

a. Stresses are usually expressed (in America) in pounds,
sometimes in tons. Thus the stress P in the preceding article is

. 1.

; .STRENGTH OF MATERIALS

v**^-* *•» 3€f''1*1 * •» . • * ?' •

1,000 pounds, or -J ton. Notice that this value has nothing to do
with the size of the cross-section on which the stress acts.

3. Kinds of Stress, (a) "When the forces acting on a body
(as a rope or rod) are such that they tend to tear it, the stress at
any cross -section is called a tension or a tensile stress. The
stresses P and Q, of Fig. 1, are tensile stresses. Stretched ropes,
loaded "tie rods" of roofs and bridges, etc., are under tensile stress.
(b.) "When the forces acting on a body (as a short post, brick,

etc.) are such that they tend to
crush it, the stress at any sec-
tion at right angles to the di-
rection of the crushing forces is
called a pressure or a compres-
sive stress. Fig. 2 («) repre-
sents a loaded post, and Fig. 2
(&) the upper and lower parts.
The upper part presses down on
B, and the lower part presses up
on A, as shown. P or Q is
the compressive stress in the
post at section C. Loaded posts,
or struts, piers, etc., are under
compressive stress.

(c.) "When the forces acting
on a body (as a rivet in a bridge

T P

Fig. 2.

joint) are such that they tend to cut or " shear " it across, the stress
at a section along which there is a tendency to cut is called a shear
or a shearing stress. This kind of stress takes its name from the
act of cutting with a pair of shears. In a material which is being
cut in this way, the stresses that are being " overcome " are shear-
ing stresses. Fig. 3 (&) represents a riveted joint, and Fig. 3 (J>)
two parts of the rivet. The forces applied to the joint are such
that A tends to slide to the left, and B to the right; then B exerts
on A a force P toward the right, and A on B a force Q toward the
left as shown. P or Q is the shearing stress in the rivet.

Tensions, Compressions and Shears are called simple stresses.
Forces may act upon a body so as to produce a combination of simple
stresses on some section ; such a combination is called a complex

STRENGTH OF MATERIALS 3

stress. The stresses in beams are usually complex. There are other
terms used to describe stress; they will be defined farther on.

4. Unit=Stress. It is often necessary to specify not merely
the amount of the entire stress which acts on an area, but also the
amount which acts on each unit of area (square inch for example).
By unit-stress is meant stress per unit area.

To find the value of a unit-stress: Divide the whole stress by
the whole area of the section on which it acts, or over which it is
distributed. Thus, let

P denote the value of the whole stress,

A the area on which it acts, and

S the value of the unit-stress; then

= AS. (l)

Strictly these formulas apply only when the stress P is uniform,

Fig. 3.

that is, when it is uniformly distributed over the area, each square
inch for example sustaining the same amount of stress. When
the stress is not uniform, that is, when the stresses on different
square inches are not equal, then P-r-A equals the average value
of the unit-stress.

5. Unit-stresses are usually expressed (in America) in
pounds per square inch, sometimes in tons per square inch. If
P and A in equation 1 are expressed in pounds and square
inches respectively, then S will be in pounds per square inch; and
if P and A are expressed in tons and square inches, S will be in
tons per square inch.

Examples. 1. Suppose that the rod sustaining the load in
Fig. 1 is 2 square inches in cross-section, and that the load weighs
1,000 pounds. What is the value of the unit-strese ?

4 STRENGTH OF MATERIALS

Here P = 1,000 pounds, A= 2 square inches; hence.
S = -— — • = 500 pounds per square inch.

2. Suppose that the rod is one-half square inch in cross -sec-
tion. What is the value of the unit-stress ?

A = -^-square inch, and, as before, P== 1,000 pounds; hence

S = 1,000-J — g- = 2,000 pounds per square inch.

Notice that one must always divide the whole stress by the area to get
the unit-stress, whether the area is greater or less than one.

6. Deformation. Whenever forces are applied to a body it
changes in size, and usually in shape also. This change of size
and shape is called deformation. Deformations are usually meas-
ured in inches; thus, if a rod is stretched 2 inches, the "elonga-
tion"= 2 inches.

7. Unit-Deformation. It is sometimes necessary to specify
not merely the value of a total deformation but its amount per
unit length of the deformed body. Deformation per unit length
of the deformed body is called unit-deformation.

To find the value of a unit-deformation : Divide the whole
deformation 1y the length over which it is distributed. Thus, if
D denotes the value of a deformation,
I the length,
s the unit-deformation, then

8=-j-, also T>=ls. (2)

Both D and I should always be expressed in the same unit.

Example. Suppose that a 4-foot rod is elongated \ inch.
What is the value of the unit-deformation?

Here D— \ inch, and 1=4: feet=48 inches;
hence «=J-j-48=-J^r inch per inch.
That is, each inch \s elongated -£v inch.

Unit-elongations are sometimes expressed in per cent. To
express a** elongation in per cent: Divide the elongation in inches
by the original length in inches, and multiply l)y 100.

8. Elasticity. Most solid bodies when deformed will regain
more or less completely their natural size and shape when the de.

STRENGTH OF MATERIALS 5

forming forces cease to act. This property of regaining size and
shape is called elasticity.

"We may classify bodies into kinds depending on the degree
of elasticity which they have, thus :

1. Perfectly elastic bodies; these will regain their orig-
inal form and size no matter how large the applied forces are if
less than breaking values. Strictly there are no such materials,
but rubber, practically, is perfectly elastic.

2. Imperfectly elastic bodies; these will fully regain their
original form and size if the applied forces are not too large, and
practically even if the loads are large but less than the breaking
value. Most of the constructive materials belong to this class.

3. Inelastic or plastic bodies; these will not regain in the
least their original form when the applied forces cease to act. Clay
and putty are good examples of this class.

9. Hooke's Law, and Elastic Limit. If a gradually increas-
ing force is applied to a perfectly elastic material, the deformation
increases proportionally to the force; that is, if P and P' denote
two values of the force (or stress), and D and D' the values of the
deformation produced by the force,

thenP:P'::D:D'.

This relation is also true for imperfectly elastic materials,
provided that the loads P and P' do not exceed a certain limit depend-
ing on the material. Beyond this limit, the deformation increases
much faster than the load; that is, if within the limit an addition
of 1,000 pounds to the load produces a stretch of 0.01 inch, beyond
the limit an equal addition produces a stretch larger and usually
much larger than 0.01 inch.

Beyond this limit of proportionality a part of the deformation
is permanent; that is, if the load is removed the body only partially
recovers its form and size. The permanent part of a deformation
is called set.

The fact that for most materials the deformation is propor-
tional to the load within certain limits, is known as Hooke's Law.
The unit-stress within which Hooke's law holds, or above which
the deformation is not proportional to the load or stress, is called
elastic limit.

6 STRENGTH OF MATERIALS

10. Ultimate Strength. By ultimate tensile, compressive,
or shearing strength of a material is meant the greatest tensile,
compressive, or shearing unit-stress which it can withstand.

As before mentioned, when a material is subjected to an in-
creasing load the deformation increases faster than the load beyond
the elastic limit, and much faster near the stage of rupture. Not
only do tension bars and compression blocks elongate and shorten
respectively, but their cross -sectional areas change also; tension
bars thin down and compression blocks "swell out" more or less.
The value of the ultimate strength for any material is ascertained
by subjecting a specimen to a gradually increasing tensile, com-
pressive, or shearing stress, as the case may be, until rupture oc-
curs, and measuring the greatest load. The breaking load divided
l)y the area of the original cross -section sustaining the stress, is the
value of the ultimate strength.

Example. Suppose that in a tension test of a wrought-iron
rod -| inch in diameter the greatest load was 12,540 pounds. "What
is the value of the ultimate strength of that grade of wrought iron?

The original area of the cross-section of the rod was
0.7854 (diameter)2=0.7854x J=0.1964 square inches; hence
the ultimate strength equals

12,540-^0.1964:=63,850 pounds per square inch, (nearly).

11. Stress- Deformation Diagram. A "test" to determine
the elastic limit, ultimate strength, and other information in re-
gard to a material is conducted by applying a gradually increasing
load until the specimen is broken, and noting the deformation cor-
responding to many values of the load. The first and second col-
umns of the following table are a record of a tension test on a steel
rod one inch in diameter. The numbers in the first column are
the values of the pull, or the loads, at which the elongation of
the specimen was measured. The elongations are given in the sec-
ond column. The numbers in the third and fourth columns are
the values of the unit-stress and unit-elongation corresponding to
the values of the load opposite to them. The numbers in the
third column were obtained from those in the first by dividing
the latter by the area of the cross-section of the rod, 0.7854
square inches. Thus,

3,930-^0.7854=5,000
7,850-^0.7854=10,000, etc.

STRENGTH OF MATERIALS

Total Pull
iu pounds, P

Deformation
in inches, D

Unit-Stress in
pounds per
square inch, S

Unit-
Deformation,
s

3930

0.00136

5000

0.00017

7850

.00280

10000

.00035

11780

.00404

15000

.00050

15710

.00538

20000

.00067

19635

.00672

25000

.00084

23560

.00805

30000

.00101

27490

.00942

35000

.00118

31415

.01080

40000

.00135

35345

.01221

45000

.00153

39270

.0144

50000

.00180

43200

.0800

55000

.0100

47125

.1622

60000

.0202

51050

.201

65000

.0251

54980

.281

70000

.0351

58910

.384

75000

.048

62832

.560

80000

.070

65200

1.600

83000

.200

The numbers in the fourth column were obtained by dividing
those in the second by the length of the specimen (or rather the
length of that part whose elongation was measured), 8 inches.
Thus,

0.00136 -^-8 = 0.00017,
.00280-^8 = .00035, etc.

Looking at the first . two columns it will be seen that the elonga-
tions are practically proportional to the loads up to the ninth load,
the increase of stretch for each increase in load being about 0.00135
inch; but beyond the ninth load the increases of stretch are much
greater. Hence the elastic limit was reached at about the ninth
load, and its value is about 45,000 pounds per square inch. The
greatest load was 65,200 pounds, and the corresponding unit-stress,
83,000 pounds per square inch, is the ultimate strength.

Nearly all the information revealed by such a test can be
well represented in a diagram called a stress-deformation diagram.
It is made as follows: Lay off the values of the unit-deformation
(fourth column) along a horizontal line, according to some con-
venient scale, from some fixed point in the line. At the points on
the horizontal line representing the various unit-elongations, lay
off perpendicular distances equal to the corresponding unit-stresses.
Then connect by a smooth curve the upper ends of all those dis-
tances, last distances laid off. Thus, for instance, the highest unit-

8 STRENGTH OF MATERIALS

elongation (0.20) laid off from o (Fig. 4) fixes the point a, and a
perpendicular distance to represent the highest unit- stress (83,000)
fixes the point 1). All the points so laid off give the curve ocb. The
part ocy within the elastic limit, is straight and nearly vertical
while the remainder is curved and more or less horizontal, especially
toward the point of rupture t>. Fig. 5 is a typical stress-defor-
mation diagram for timber, cast iron, wrought iron, soft and hard
steel, in tension and compression.

12. Working Stress and Strength, and Factor of Safety.
The greatest unit-stress in any part of a structure when it is sus-

taining its loads is called the
working stress of that part. If
it is under tension, compression
and shearing stresses, then the
corresponding highest unit-
stresses in it are called its work-
ing stress in tension, in com-
pression, and in shear respect-
ively; that is, we speak of as
~ """ many working stresses as it has

kinds of stress.

By working strength of a material to be used for a certain
purpose is meant the highest unit- stress to which the material
ought to be subjected when so used. Each material has a working
strength for tension, for compression, and for shear, and they are in
general different.

By factor of safety is meant the ratio of the ultimate strength
of a material to its working stress or strength. Thus, if

O O '

Su denotes ultimate strength,
Sw denotes working stress or strength, and
f denotes factor of safety, then

When a structure which has to stand certain loads is about
to be designed, it is necessary to select working strengths or fac-
tors of safety for the materials to be used. Often the selection is
a matter of great importance, and can be wisely performed only
by an experienced engineer, for this is a matter where hard -and-

STRENGTH OF MATERIALS 9

fast rules should not govern but rather the judgment of the expert.
But there are certain principles to be used as guides in making a
selection, chief among which are:

1. The working strength should be considerably below the
elastic limit. (Then the deformations will bft small and not per-
manent.)

Fig. 5. (After Johnson.)

2. The working strength should be smaller for parts of a
structure sustaining varying loads than for those whose loads are
steady. (Actual experiments have disclosed the fact that the
strength of a specimen depends on the kind of load put upon it,
and that in a general way it is less the less steady the load is.)

3. The working strength must be taken low for non -uniform
material, where poor workmanship may be expected, when the

STRENGTH OF MATERIALS

loads are uncertain, etc. Principles 1 and 2 have been reduced
to figures or formulas for many particular cases, but the third must
remain a subject for display of judgment, and even good guessing
in many cases.

The following is a table of factors of safety* which will be
used in the problems:

Factors of Safety.

Materials.

For steady
stress.
(Buildings.)

For varying
stress.
(Bridges.)

For shocks.
(Machines.)

Timber
Brick and stone
Cast iron
Wrought iron
Steel

8
15
6
4
5

10
25
15
6

15
30
20
10
15

They must be regarded as average values and are not to be
adopted in every case in practice.

Examples. 1. A wrought -iron rod 1 inch in diameter sus-
tains a load of 30,000 pounds. "What is its working stress? If
its ultimate strength is 50rOOO pounds per square inch, what is
its factor of safety ?

The area of the cross- section of the rod equals 0.7854 X (diam-
eter ) 2— 0.7854 X 12=0.7854 square inches. Since the whole stress
on the cross-section is 30,000 pounds, equation 1 gives for the
unit working stress

30 000
S = ~ VGKA — 38,197 pounds per square inch.

Equation 3 gives for factor of safety

50,000

/-a

__ -i o

38,197

2. How large a steel bar or rod is needed to sustain a steady
pull of 100,000 pounds if the ultimate strength of the material is
65,000 pounds ?

The load being steady, we use a factor of safety of 5 (see table
above); hence the working strength to be used (see equation 3) is

S = — ^— = 13,000 pounds per square inch.

The proper area of the cross -section of the rod can now be com-
puted from equation 1 thus:
*Taken from Merriman's "Mechanics of Materials/'

STRENGTH OF MATERIALS 11

P 100,000
A = g- - = -13 ' = 7.692 square inches.

A bar 2x4 inches in cross-section would be a little stronger
than necessary. To find the diameter (d) of a round rod of suffi-
cient strength, we write 0.7854 d2 = 7.692, and solve the equation
fore?/ thus:

7 fiQ2
d2=^j^= 9.794, or rf = 3,129 inches.

3. How large a steady load can a short timber post safely sus-
tain if it is 10x10 inches in cross-section and its ultimate com-
pressive strength is 10,000 pounds per square inch ?

According to the table (page 12) the proper factor of safety is
8, and hence the working strength according to equation 3 is

10,000
S = — -K — = 1,250 pounds per square inch.

The area of the cross-section is 100 square inches; hence the safe
load (see equation 1) is

P = 100 X 1,250 = 125,000 pounds.

4. When a hole is punched through a plate the shearing
strength of the material has to be overcome. If the ultimate shear-

ing strength is 50,000 pounds per square inch, the thickness of the
plate -J inch, and the diameter of the hole f inch, what is the value
of the force to be overcome ?

The area shorn is that of the cylindrical surface of the hole
or the metal punched out^ that is

3.1416 X diameter X thickness = 3.1416 X f X J = 1.178 sq. in.
Hence, by equation 1, the total shearing strength or resistance
to punching is

P = 1.178 X 50,000 = 58,900 pounds.

STRENGTH OF MATERIALS UNDER SIMPLE STRESS.

13. Materials in Tension. Practically the only materials
used extensively under tension are timber, wrought iron and steel,
and to some extent cast iron.

14. Timber. A successful tension test of wood is difficult,
as the specimen usually crushes at the ends when held in the test-
ing machine, splits, or fails otherwise than as desired. Hence the

12 STRENGTH OF MATERIALS

tensile strengths of woods are not well known, but the following
may be taken as approximate average values of the ultimate
strengths of the woods named, when "dry out of doors."

Hemlock, 7,000 pounds per square inch.

White pine, 8,000 " " .

Yellow pine, long leaf, 12,000 "

" " , short leaf, 10,000 "

Douglas spruce, 10,000 "

White oak, 12,000

Red oak, 9,000

15. Wrought Iron. The process of the manufacture of
wrought iron gives it a "grain," and its tensile strengths along and
across the grain are unequal, the latter being about three-fourths
of the former. The ultimate tensile strength of wrought iron
along the grain varies from 45,000 to 55,000 pounds per square
inch. Strength along the grain is meant when not otherwise
Btated.

The strength depends on the size of the piece, it being greater
for small than for large rods or bars, and also for thin than for
thick plates. The elastic limit varies from 25,000 to 40,000
pounds per square inch, depending on the size of the bar or plate
even more than the ultimate strength. Wrought iron is very
ductile, a specimen tested in tension to destruction elongating from
5 to 25 per cent of its length.

16. Steel. Steel has more or less of a grain but is practically
^f the same strength in all directions. To suit different purposes,
steel is made of various grades, chief among which may be men-
tioned rivet steel, sheet steel (for boilers), medium steel (for
bridges and buildings), rail steel, tool and spring steel. In general,
these grades of steel are hard and strong in the order named, the
ultimate tensile strength ranging from about 50,000 to 160,000
pounds per square inch.

There are several grades of structural steel, which may be
described as follows:*
1. Rivet steel:

Ultimate tensile strength, 48,000 to 58,000 pounds per square inch.
Elastic limit, not less than one-half the ultimate strength.
Elongation, 26 per cent.
Bends 180 degrees flat on itself without fracture.

*Taken from " Manufacturer's Standard Specifications."

STRENGTH OF MATERIALS 13

2. Soft steel:

Ultimate tensile strength, 52,000 to 62,000 pounds per square inch.
Elastic limit, not less than one-half the ultimate strength.
Elongation, 25 per cent.
Bends 180 degrees flat on itself.

3. Medium steel:

Ultimate tensile strength, 60,000 to 70,000 pounds per square inch.
Elastic limit, not less than one-half the ultimate strength. •
Elongation, 22 per cent.

Bends 180 degrees to a diameter equal to the thickness of the
specimen without fracture.

17. Cast Iron. As in the case of steel, there are many
grades of cast iron. The grades are riot the same for all localities
or districts, but they are based on the appearance of the fractures,
which vary from coarse dark grey to fine silvery white.

The ultimate tensile strength does not vary uniformly with
the grades but depends for the most part on the percentage of
"combined carbon" present in the iron. This strength varies from
15,000 to 35,000 pounds per square inch, 20,000 being a fair
average.

Cast iron has no well-defined elastic limit (see curve for cast
iron, Fig. 5). Its ultimate elongation is about one per cent.

EXAMPLES FOR PRACTICE.

1. A steel wire is one-eighth inch in diameter, and the ulti-
mate tensile strength of the material is 150,000 pounds per square
inch. How large is its breaking load ? Ans. 1,845 pounds.

2. A wrought-iron rod (ultimate tensile strength 50,000
pounds per square inch) is 2 inches in diameter. How large a
steady pull can it safely bear ? Ans. 39,270 pounds.

18. Materials in Compression. Unlike the tensile, the
compressive strength of a specimen or structural part depends on
its dimension in the direction in which the load is applied, for,
in compression, a long bar or rod is weaker than a short one. At
present we refer only to the strength of short pieces such as do
not bend under the load, the longer ones (columns) being dis-
cussed farther on.

Different materials break or fail under compression, in two
very different ways:

1. Ductile materials (structural steel, wrought iron, etc.),

14 STRENGTH OF MATERIALS

and wood compressed across the grain, do not fail by breaking Into
two distinct parts as in tension, but the former bulge out and
flatten under great loads, while wood splits and mashes down.
There is no particular point or instant of failure under increasing
loads, and such materials have no definite ultimate strength in
compression.

2. Brittle materials (brick, stone, hard steel, cast iron, etc.),
and wood compressed along the grain, do not mash gradually, but
fail suddenly and have a definite ultimate strength in compression.
Although the surfaces of fracture are always much inclined to the
direction in which the load is applied (about 45 degrees), the ulti.
mate strength is computed by dividing the total breaking load by
the cross -sectional area of the specimen.

The principal materials used under compression in structural
work are timber, wrought iron, steel, cast iron, brick and stone.

19. Timber. As before noted, timber has no definite ulti-
mate compressive strength across the grain. The U. S. Forestry
Division has adopted certain amounts of compressive deformation
as marking stages of failure. Three per cent compression is
regarded as "a working limit allowable," and fifteen per cent as
"an extreme limit, or as failure." The following (except the first)
are values for compressive strength from the Forestry Division
Reports, all in pounds per square inch:

Ultimate strength 3£ Compression
along the grain. across the grain

Hemlock 6,000

White pine 5,400 700

Long-leaf yellow pine 8,000 1,260

Short-leaf yellow pine 6,500 1,050

Douglas spruce 5,700 800

White oak 8,500 2,200

Red oak 7,200 2,300 .

20. Wrought Iron. The elastic limit of wrought iron, as be-
fore noted, depends very much upon the size of the bars or plate, it
being greater for small bars and thin plates. Its value for com-
pression is practically the same as for tension, 25,000 to 40,000
pounds per square inch.

21. Steel. The hard steels have the highest compressive
strength; there is a recorded value of nearly 400,000 pounds per
square inch, but 150,000 is probably a fair average.

STRENGTH OF MATERIALS 15

The elastic limit in compression is practically the same as in
tension, which is about 60 per cent of the ultimate tensile strength,
or, for structural steel, about 25,000 to 42,000 pounds per square

inch.

22. Cast Iron. This is a very strong material in compres-
sion, in which way, principally, it is used structurally. Its ulti-
mate strength depends much on the proportion of "combined car-
bon" and silicon present, and varies from 50,000 to 200,000 pounds
per square inch, 90,000 being a fair average. As in tension,
there is no well-defined elastic limit in compression (see curve for
cast iron, Fig. 5).

23. Brick. The ultimate strengths are as various as the
kinds and makes of brick. For soft brick, the ultimate strength
is as low as 500 pounds per square inch, and for pressed brick it
varies from 4,000 to 20,000 pounds per square inch, 8,000 to
10,000 being a fair average. The ultimate strength of good pav-
ing brick is still higher, its average value being from 12,000 to
15,000 pounds per square inch.

24. Stone. Sandstone, limestone and granite are the
principal building stones. Their ultimate strengths in pounds
per square inch are about as follows:

Sandstone,* 5,000 to 16,000, average 8,000.
Limestone,* 8,000 " 16,000, " 10,000.
. Granite, 14,000 " 24,000, " 16,000.
Compression at right angles to the "bed" of the stone.

EXAMPLES FOR PRACTICE.

1. A limestone 12 X 12 inches on its bed is used as a pier
cap, and bears a load of 120,000 pounds. What is its factor of
safety ? Ans. 12.

2. How large a post (short) is needed to sustain a steady
load of 100,000 pounds if the ultimate compressive strength of
the wood is 10,000 pounds per square inch ? Ans. 8 X 10 inches.

25. Materials in Shear. The principal materials used under
shearing stress are timber, wrought iron, steel and cast iron.
Partly on account of the difficulty of determining shearing
strengths, these are not well known.

26. Timber. The ultimate shearing strengths of the more
important woods along the grain are about as follows:

16 STRENGTH OF MATERIALS

•

Hemlock, 300 pounds per square inch.

White pine, 400 " "

Long-leaf yellow pine, 850 " "

Short-leaf " " 775 " "

Douglas spruce, 500 " "

White oak, 1,000 • " "

Red oak, 1,100 " "

Wood rarely fails by shearing across the grain. Its ultimate

Fig. 6 a. Fig. 6 b.

shearing strength in that direction is probably four or five timsa
the values above given.

27. Metals. The ultimate shearing strength of wrought
iron, steel, and cast iron is about 80 per cent of their respective
ultimate tensile strengths.

EXAMPLES FOR PRACTICE.

1. How large a pressure P (Fig. 6 a) exerted on the shaded
area can the timber stand before it will shear off on the surface
abed) if ab = 6 inches and "bo = 10 inches, and the ultimate shear-
ing strength of the timber is 400 pounds per square inch ?

Ans. 24,000 pounds.

2. When a bolt is under tension, there is a tendency to tear
the bolt and to "strip" or shear off the head. The shorn area
would be the surface of the cylindrical hole left in the head.
Compute the tensile and shearing unit-stresses when P (Fig. 6 £)
equals 30,000 pounds, d — 2 inches, and t — 3 inches.

j Tensile unit-stress, 9,550 pounds per square inch.
" ( Shearing unit-stress, 1,591 pounds per square inch.

REACTIONS OF SUPPORTS.

28. Moment of a Force. By moment of a force with re-
epect to a point is meant its tendency to produce rotation about
that point. Evidently the tendency depends on the magnitude of
the force and on the perpendicular distance of the line of action
of the force from the point : the greater the force and the per-
pendicular distance, the greater the tendency; hence the moment

STRENGTH OF MATERIALS

of a force with respect to a point equals the product of the force
and the perpendicular distance from the force to the point.

The point with respect to which the moment of one or more
forces is taken is called an origin or center of moments, and the
perpendicular distance from an origin of moments to the line of
action of a force is called the arm of the force with respect to
that origin. Thus, if F1 and F2 (Fig. 7) are forces, their arms
with respect to O' are «/ and a2' respectively, and their moments
are F^'i and F2<r'2. "With respect to O" their arms are #/' and a"
respectively, and their moments are F^/' and F2«2".

If the force is expressed in pounds and its arm in feet, the
moment is in foot-pounds; if the force is in pounds and the arm
in inches, the moment is in inch-pounds.

29. A sign is given to the moment of a force for conven-
ience; the rule used herein is as follows: The moment of a
force about a point is positive or negative according as it tends
to turn the hody about that point in the clockwise or counter-
clockwise direction*.

Thus the moment (Fig. 7)

of Fj about O' is negative, about O" positive;
" F2 " O' « « , about O" negative.

30. Principle of Moments. In general, a single force of
proper magnitude and line of ac-
tion can balance any number of

forces. That single force is called
the equilibrant of the forces, and
the single force that would balance
the equilibrant is called the result-
ant of the forces. Or, otherwise
stated, the resultant of any num-
ber of forces is a force which pro-
duces the same effect. It can be
proved that — The algebraic sum
of the moments of any number
of forces with respect to a point,
equals the moment of their re- Y\S. 1.

sultant about that point.

*By clockwise direction is meant that in which the hands of a clock
rotate; and by counter-clockwise, the opposite direction.

18 STRENGTH OF MATERIALS

This is a useful principle and is called "principle of moments."

31. All the forces acting upon a body which is at rest are

said to be balanced or in equilibrium. "No force is required to

balance such forces and hence their equilibrant and resultant are

zero.

Since their resultant is zero, the algebraic sum of the mom-

looolbs. aooolbs. aooolbs. loooltos.

A r^ is

5 v

u lC -5 u
Fig. 8.

ents of any number of forces which are balanced or in equilib-
. rium equals zero.

This is known as the principle of moments for forces in
equilibrium; for brevity we shall call it also "the principle of
moments."

The principle is easily verified in a simple case. Thus, let
AB (Fig. 8) be a beam resting on supports at C and F. It is
evident from the symmetry of the loading that each reaction
equals one-half of the whole load, that is, \ of 6,000=3,000
pounds. (We neglect the weight of the beam for simplicity.)

With respect to C, for example, the moments of the forces
are, taking them in order from the left:

—1,000 X 4 = — 4,000 foot-pounds

3,000 X 0= 0 "

2,000 X 2= 4,000

2,000 X 14 = 28,000
—3,000 X 16 = — 48,000 "

1,000 X 20 = 20,000

The algebraic sum of these moments is seen to equal zero.
Again, with respect to B the moments are:

— 1,000 X 24 = — 24,000 foot-pounds

3,000 X 20 = 60,000
— 2,000. X 18 = -— 36,000
— 2,000 X 6 = — 12,000

3,000 X 4= 12,000

1,000 X 0= 0

The sum of these moments also equals zero. In fact, no matter

STRENGTH OF MATERIALS 19

where the center of moments is taken, it will be found in this and
any other balanced system of forces that the algebraic sum of their
moments equals zero. The chief .use that we shall make of this
principle is in finding the supporting forces of loaded beams.

32. Kinds of Beams. A cantilever 'beam is one resting on
one support or fixed at one end, as in a wall, the other end being
free.

A simple beam is one resting on two supports.

A restrained beam is one fixed at both ends; a beam fixed at
one end and resting on a support at the other is said to be re-
strained at the fixed end and simply supported at the other.

A continuous beam is one resting on more than two supports.

33. Determination of Reactions on Beams. The forces which
the supports exert on a beam, that is, the "supporting forces," are
called reactions. We shall deal chiefly with simple beams. The
reaction on a cantilever beam supported at one point evidently
equals the total load on the beam.

When the loads on a horizontal beam are all vertical (and

looolbs. 2000 Ibs. aooolbs.

L, ,1

t.,1 J, ^1 J

K — 1 *

- 1 -

T B

'R!

O i-

D
6» *

J ^.

t*.. £-•*•

Fig. 9.

^ -

^ i

this is the usual case), the supporting forces are also vertical and
the sum of the reactions equals the sum of the loads. This prin-
ciple is sometimes useful in determining reactions, but in the case
of simple beams the principle of moments is sufficient. The gen-
eral method of determining reactions is as follows:

1. Write out two equations of moments for all the forces
(loads and reactions) acting on the beam with origins of moments
at the supports.

2. Solve the equations for the reactions.

3. As a check, try if the sum of the reactions equals the
sum of the loads.

Examples. 1. Fig. 9 represents a beam supported at its
ends and sustaining three loads. We wish to find the reactions
due to these loads.

20 STRENGTH OF MATERIALS

Let the reactions be denoted by Rt and R2 as shown; then
the moment equations are:
For origin at A,

1,000 X 1 + 2,000 X 6 + 3,000 X 8— R2 X 10 = 0.
For origin at E,

2100lbS. 3600 llOS.

La,_ 6. 1

I [B JC .

Fig. 10.

Ri x 10—1,000 X 9—2,000 X 4—3,000 X 2 = 0.
The first equation reduces to

10 R2 = 1,000+12,000+24,000 = 37,000; or
R2= 3,700 pounds.
The second equation reduces to

10 E1= 9,000+8,000+6,000 = 23,000; or
B1== 2,300 pounds.

The sum of the loads is 6,000 pounds and the sum of the reactions
is the same; hence the computation is correct.

2. Fig. 10 represents a beam supported at B and D (that is,
it has overhanging ends) and sustaining three loads as shown. We
wish to determine the reactions due to the loads.

Let Rj and R2 denote the reactions as shown ; then the moment
equations are:
For origin at B,

-2,100x2+0+3,600x6— R2X 14+ 1,600x18 = 0.
For origin at D,

-2,100x16+^x14—3,600x8 + 0+1,600x4 = 0.
The first equation reduces to

14 Ea= -4,200+21,600 + 28,800 = 46,200; or
R2 = 3,300 pounds.
The second equation reduces to

14 R1== 33,600+28,800-6,400 = 56,000; or
E1= 4,000 pounds.

The sum of the loads equals 7,300 pounds and the sum of the
reactions is the same; hence the computation checks.

3. What are the total reactions in example 1 if the beam
weighs 400 pounds ?

STRENGTH OF MATERIALS 21

(1.) Since we already know the reactions due to the loads
(2,300 and 3,700 pounds at the left and right ends respectively
(see illustration 1 above), we need only to compute the reactions
due to the weight of the beam and add. Evidently the reactions
due to the weight equal 200 pounds each; hence the

left reaction =2,300+200—2,500 pounds, and the
-right « = 3,700+ 200— 3,900 « .

(2.) Or, we might compute the reactions due to the loads
and weight of the beam together and directly. In figuring the
moment due to the weight of the beam, we imagine the weight
as concentrated at the middle of the beam ; then its moments with
respect to the left and right supports are (400 X 5) and — (400 X 5)
respectively. The moment equations for origins at A and E are
like those of illustration 1 except that they contain one more
term, the moment due to the weight ; thus they are respectively :

1,000x1 + 2,000x6 + 3,000x8— R2X 10 +400X5-0,

E! X 10—1,000 X 9—2,000 X 4—3,000 X 2—400 X 5 -0.
The first one reduces to

10 E2= 39,000, or R, = 3,900 pounds;
and the second to

10 .E1= 25,000, or E1== 2,500 pounds.

4. What are the total reactions in example 2 if the beam
weighs 42 pounds per foot ?

As in example 3, we might compute the reactions due to the
weight and then add them to the corresponding reactions due to
the loads (already found in example 2), but we shall determine
the total reactions due to load and weight directly.

The beam being 20 feet long, its weight is 42x20, or 840
pounds. Since the middle of the beam is 8 feet from the left and
6 feet from the right support, the moments of the weight with
to the left and right supports are respectively:

840X8 = 6,720, and— 840x6 — —5,040 foot-pounds.
The moment equations for all the forces applied to the beam
for origins at B and D are like those in example 2, with an addi-
tional term, the moment of the weight; "they are respectively:
— 2,100x2 + 0+3,600x6— R2 X 14+ 1,600 X 18 + 6,720 = 0,
—2,100 X 16 + Rj X 14—3,600 X 8 + 0 + 1,600 X 4—5,040 = 0.

22 STRENGTH OF MATERIALS

The first equation reduces to

14 R2=52,920, or R2=3,780 pounds,
and the second to

14 R^ 61,040, or R^ 4,360 pounds.

The sum of the loads and weight of beam is 8,140 pounds;
and since the sum of the reactions is the same, the computation
checks.

EXAMPLES FOR PRACTICE.

1. AB (Fig. 11) represents a simple beam supported at its
ends. Compute the reactions, neglecting the weight of the beam,

. j Right reaction = 1,443.75 pounds.
'' J Left reaction = 1,556.25 pounds.

eoolbs. 900 Ibs. soolbs. looolbs.

k— 2'— * 4-

Fig. 11.

2. Solve example 1 taking into account the weight of the
beam, which suppose to be 400 pounds.

. ( Right reaction = 1,643.75 pounds.
| Left reaction = 1,756.25 pounds.

3. Fig. 12 represents a simple beam weighing 800 pounds
supported at A and B, and sustaining three loads as shown.
"What are the reactions ?

. j Right reaction = 2,014.28 pounds.
' ( Left reaction = 4,785.72 pounds.

2000 Ibs. 1000 Ibs. 3000 Ibs.

— •' — t »'— t
*

Fig. 12.

4. Suppose that in example 3 the beam also sustains a uni-
formly distributed load (as a floor) over its entire length, of 500
pounds per foot. Compute the reactions due to all the loads and
the weight of the beam.

. j Right reaction = 4,871.43 pounds.
QS* ( Left reaction =±= 11,928.57 pounds.

STRENGTH OF MATERIALS 23

EXTERNAL SHEAR AND BENDING MOMENT.

On almost every cross-section of a loaded beam there are
three kinds of stress, namely tension, compression and shear. The
first two are often called fibre stresses because they act along the
real fibres of a wooden beam or the imaginary ones of which we
may suppose iron and steel beams composed. Before taking up
the subject of these stresses in beams it is desirable to study certain
quantities relating to the loads, and on which the stresses in a
beam depend. These quantities are called external shear and
bending moment, and will now be discussed.

34. External Shear. By external shear at (or for) any sec-
tion of a loaded beam is meant the algebraic sum of all the loads
(including weight of beam) and reactions on either side of the
section. This sum is called external shear because, as is shown
later, it equals the shearing stress (internal) at the section. For
brevity, we shall often say simply "shear" when external shear is
meant.

35, Rule of Signs. In computing external shears, it is cus-
tomary to give the plus sign to the reactions and the minus sign
to the loads. But in order to get the same sign for the external
shear whether computed from the right or left, we change the sign
of the sum when computed from the loads and reactions to the
right. Thus for section a of the beam in Fig. 8 the algebraic sum is,
when computed from the left,

-1,000 + 3,000= +2,000 pounds;
and when computed from the right,

-1,000+3,000-2,000-2,000 = -2,000 pounds.
The external shear at section a is +2,000 pounds.

Again, for section b the algebraic sum is,
when computed from the left,

-1,000 + 3,000-2,000-2,000 + 3,000 = + 1,000 pounds;
and when computed from the right, -1,000 pounds.

The external shear at the section is + 1,000 pounds.

It is usually convenient to compute the shear at a section
from the forces to the right or left according as there are fewer
forces (loads and reactions) on the right or left sides of the
section.

24 STRENGTH OF MATERIALS

36. Units for Shears. It is customary to express external
shears in pounds, but any other unit for expressing force and
weight (as the ton) may be used.

37. Notation. We shall .use Y to stand for external shear at
any section, and the shear at a particular section will be denoted
by that letter subscripted; thus Y1? Y2, etc., stand for the shears
at sections one, two, etc., feet from the left end of a beam.

The shear has different values just to the left and right of a
support or concentrated load. We shall denote such values by Y'
and Y"; thus Y5' and Y5" denote the values of the shear at sec-
tions a little less and a little more than 5 feet from the left end
respectively.

Examples. 1. Compute the shears for sections one foot
apart in the beam represented in Fig. 9, neglecting the weight of
the beam. (The right and left reactions are 3,700 and 2,300
pounds respectively; see example 1, Art. 33.)

All the following values of the shear are computed from the
left. The shear just to the right of the left support is denoted by
Y0", and Y0" = 2,300 pounds. The shear just to the left of B is
denoted by Y/, and since the only force to the left of the section
is the left reaction, Y/ = 2,300 pounds. The shear just to the
right of B is denoted by Y/', and since the only forces to the left
of this section are the left reaction and the 1,000-pound load,
Y/' = 2,300 - 1,000 = 1,300 pounds. To the left oj all sections
between B and C, there are but two forces, the left reaction and
the 1,000-pound load; hence the shear at any of those sections
equals 2,300-1,000^1,300 pounds, or

Y2 = Y3 = Y4 = Y5 = Y6'= 1,300 pounds.

The shear just to the right of 0 is denoted by Y6"; and since the
forces to the left of that section are the left reaction and the
1,000- and 2,000-pound loads,

Y6" == 2,300 - 1,000 - 2,00l « - 700 pounds.

Without further explanation, thb student should understand
that

Y7 = + 2,300 - 1,000 - 2,000 === - 700 pounds,

Y; =-700,

Y8" = + 2,300 - 1,000 - 2,000 - 3,000 = - 3,700,

Y9 = V10'=- 3,700,

Y10" = + 2.300 - 1,000 - 2,000 - 3,000 + 3,700 = 0

STRENGTH OF MATERIALS 25

2. A simple beam 10 feet long, and supported at each end,
weighs 400 pounds, and bears a uniformly distributed load of
1,600 pounds. Compute the shears for sections two feet apart.

Evidently each reaction equals one-half the sum of the load
and weight of the beam, that is, \ (1,600+400) =1,000 pounds.
To the left of a section 2 feet from the left end, the forces acting
on the beam consist of the left reaction, the load on that part of
the beam, and the weight of that part ; then since the load and
weight of the beam per foot equal 200 pounds,

Y2= 1,000-200 X 2 = 600 pounds.

To the left of a section four feet from the left end, the forces
are the left reaction, the load on that part of the beam, and the
weight ; hence

V \= 1,000-200 X 4 = 200 pounds.

Without further explanation, the student should see that
Y6 = 1,000-200 X 6 =-200 pounds,
Y8 = 1,000-200 X 8 = -600 pounds,
V10' = 1,000-200 X 10 = -1,000 pounds,
V10"= 1,000-200x10+1,000 = 0.

3. Compute the values of the shear in example 1, taking
into account the weight of the beam (400 pounds). (The right
and left reactions are then 3,900 and 2,500 pounds respectively;
see example 3, Art. 33.)

We proceed just as in example 1, except that in each compu-
tation we include the weight of the beam to the left of the section
(or to the right when computing from forces to the right). The
weight of the beam being 40 pounds per foot, then (computing
from the left)

Vo"=+ 2,500 pounds,

V/ =+2,500-40 =+2,460,

V/' =+2,500-40-1,000= + 1,460,

V2 =+2,500-1,000-40x2 =+1,420,

Y3 = f 2,500-1,000-40 X 3 = + 1,380,

V.4 =+ 2,500-1,000-40 X 4 = + 1,340,

Y5 =+2,500-1,000-40x5 = +1,300,

V6' =+ 2,500-1,000-40 X 6 = + 1,260,

Y 6" =+ 2,500-1,000-40 X 6-2,000 = -740,

V7 =+2,500-1,000-2,000-40 X 7 = -780,

26 STRENGTH OF MATERIALS

V8' = + 2,500-1,000-2,000-40 X 8 = -820,
V8" = + 2,500-1,000-2,000-40x8-3,000 =-3,820,
Yg = + 2,500-1,000-2,000-3,000-40 X 9 = -3,860,
V'io = + 2,500-1,000-2,000-3,000-40 X 10 = -3,900,
Y"10= + 2,500-1,000-2,000-3,000-40x10 + 3,900=0.

Computing from the right, we find, as before, that

V7 =-(3,900-3,000-40 X 3) =-780 pounds,
Y/ =_(35900-3,000-40X2)=-820,
V8" =-(3,900-40 X 2)^-3,820,
etc., etc.

EXAMPLES FOR PRACTICE.

1. Compute the values of the shear for sections of the beam
represented in Fig. 10, neglecting the weight of the beam. (The
right and left reactions are 3,300 and 4,000 pounds respectively;
see example 2, Art. 33.)

rV, =V2'=- 2,100 pounds,

Ans J Y*" =V.=V4=VB=V6=VT=V'8= + 1,900,

' 1 V8" =Y9=Y10=Y11=Y12=Y13=YU=Y15=Y'16=-1,700,
L V"16 =Y17=Y18-Y19=Y'20= + 1,600.

2. Solve the preceding example, taking into accoun-t the
weight of the beam, 42 pounds per foot. (The right and left
reactions are 3,780 and 4,360 pounds respectively; see example 4,
Art. 33.)

Y0" = - 2,100 Ibs. Y7 = + 1,966 Ibs. Yu =- 1,928 Ibs.

Y1 =-2,142 Y8' = + 1,924 Y15 =- 1,970

Y2' = - 2,184 Y8" = - 1,676 Y16' = - 2,012

Y2" = + 2,176 Y9 =-1,718 V16"= +1,768

Y3 = + 2,134 Y10 = - 1,760 Y17 = + 1,726

y4 = + 2,092 Yn =-1,802 YI8 = + 1,684

Y5 = + 2,050 Y12=- 1,844 Y19 = + 1,642

ye = + 2,008 Y13 = - 1,886 VM' = + 1,600

3. Compute the values of the shear at sections one foot apart
in the beam of Fig." 11, neglecting the weight. (The right and
left reactions are 1,444 and 1,556 pounds respectively; see example
1, Art. 33.)

Ans.

STRENGTH OP MATERIALS 27

V0" =V1=V/= + 1,556 pounds,
V," =¥,=¥,=¥,=¥;= +956,
V6" =V/= + 56,
V," =V8=V9=V10=V11=V12=V13'=-444,
[ V13"=V14=V15=V16'=-1,444.

4. Compute the vertical shear at sections one foot apart ID
the beam of Fig. 12, taking into account the weight of the beam,
800 pounds, and a distributed load of 500 pounds per foot. (The
right and left reactions are 4,870 and 11,930 pounds respectively;
see examples 3 and 4, Art. 33.)

Y0 = 0 Y7 = + 6,150 Ibs. Y15 =+ 830 Ibs,

Vx' = - 540 Ibs. V8' = +5,610 Y16 = + 290
VY'=- 2,540 V8" =+4,610 Y17' = - 250
Y2 = _ 3,080 Y9 =+4,070 Y17" = -3,250

Ans. -<

Y3 = - 3,620 Y10 = + 3,530 Y18 = -3,790

Y4 = - 4,160 Yn = +2,990 Y19 = - 4,330

Y5 = - 4,700 Y12 = +2,450 Y20' = - 4,870

Y6' = - 5,240 Y13 = + 1,910 Y20"= 0

Y6"=+ 6,690 VM= + 1,370

38. Shear Diagrams. The way in which the external shear
varies from section to section in a beam can be well represented
by means of a diagram called a shear diagram. To construct
such a diagram for any loaded beam,

1. Lay off a line equal (by some scale) to the length of
the beam, and mark the positions of the supports and the loads.
(This is called a "base-line.")

2. Draw a line such that the distance of any point of it
from the base equals (by some scale) the shear at the correspond-
ing section of the beam, and so that the line is above the base
where the shear is positive, and below it where negative. (This is
called a shear line, and the distance from a point of it to the
base is called the "ordinate" from the base to the shear line at
that point.)

We shall explain these diagrams further by means of illus-
trative examples.

Examples. 1. It is required to construct the shear diagram
for the beam represented in Fig. 13, a (a copy of Fig. 9).

28 STRENGTH OF MATERIALS

Lay off A'E' (Fig. 13, &) to represent the beam, and mark the
positions of the. loads B', C' and D'. In example 1, Art. 37, we
computed the values of the shear at sections one foot apart; hence
we lay off ordinates at points on A'E' one foot apart, to represent
those shears.

Use a scale of 4,000 pounds to one inch. Since the shear for
any section in AB is 2,300 pounds, we draw a line ab parallel
to the base 0.575 inch (2,300 ^-4,000) therefrom; this is the shear
line for the portion AB. Since the shear for any section in BC
equals 1,300 pounds, we draw a line Vc parallel to the base and

looolbs.

2000 Ibs. 3000 Ibs.

ra~T~" o — * — * — » —

2 M

JB Tc D
i i i
*. b

111 I I i r \,

A B' c1 Illllllllillllllllil

c1 d
ScdJe: in- 4-000 Itos.

d1 ,

'a.

Fig. 13.

0.325 inch (1,300-f- 4,000) therefrom; this is the shear line for the
portion BC. Since the shear for any section in CD is -700
pounds, we draw a line cd below the base and 0.175 inch
(700-^-4,000) therefrom; this is the shear line for the portion
CD. Since the shear for any section in DE equals -3,700 Ibs., we
draw a liued'e below the base and 0.925 inch (3,700^-4,000) there-
from; this is the shear line for the portion DE. Fig. 13, £, is the
required shear diagram.

2. It is required to construct the shear diagram for the
beam of Fig. 14, a (a copy of Fig. 9), taking into account the
weight of the beam, 400 pounds.

The values of the shear for sections one foot apart were com-
puted in example 3, Art. 37, so we have only to erect ordinates at
the various noints on a base line A'E' (Tier. 14. l>\ eaual to those

STRENGTH OF MATERIALS

values. We shall use the same scale as in the preceding illustra-
tion, 4,000 pounds to an inch. Then the lengths of the ordinates
corresponding to the values of the shear (see example 3, Art. 37}
are respectively:

2,500-^4,000=0.625 inch

2,460-^4,000=0.615 "

1,460-^4,000=0.365 «

etc. etc.

Laying these ordinates off from the base (upwards or downwards
according as they correspond to positive or negative shears), we
get ob, J'tf, c'd, and d'e as the shear lines.

iooolbs. aooolfos. 3000 Ibs.

, ,

c *° :c 2.' ii

r-1 -

i i
i i

B
b

c ID

c !
!D'

3. It is required to construct the shear diagram for the
cantilever beam represented in Fig. 15, a, neglecting the weight
of the beam.

The value of the shear for any section in AB is — 500 pounds ;
for any section in BC, -1,500 pounds; and for any section in
CD, - 3,500 pounds. Hence the shear lines are ab, b'c, c'd. The
scale being 5,000 pounds to an inch,

A'« = 500-^5,000 = 0.1 inch,

B'£' = 1,500-^5,000 = 0.3 «

CV = 3,500-^5,000 = 0.7 «

The shear lines are all below the base because all the values of the
shear are negative.

STRENGTH OF MATERIALS

4. Suppose that the cantilever of the preceding illustration
sustains also a uniform load of 200 pounds per foot (see Fig. 16, a).
Construct a shear diagram.

2000 Ibs.

First, we compute the values of the shear at several sections.
Thus V0" ==- 500 pounds,

V, =-500 -200= -700,

Y2' =_500-200x2^-900,
Y2" =_ 500 - 200 X 2 - 1,000^-1,900,
Y3 =- 500 - 1,000 - 200 X 3=-.2,100,
Y4 ==- 500 - 1,000 - 200x4^-2,300,
Y5' =-500 - 1,000 - 200x5^-2,500,
Y5" =- 500 - 1,000 - 200X5 - 2,000—4,500,
V6 =- 500 - 1,000 - 2,000 - 200x6^-4,700,
Y7 =_ 500 - 1,000 - 2,000 - 200 X 7^-4,900,
Y8 =- 500 - 1,000 - 2,000 - 200 X 8=-5,100,
Y9 = -500 - 1,000 - 2,000 - 200x9^-5,300.
The values, being negative, should be plotted downward. To a
scale of 5,000 pounds to the inch they give the shear lines #5, b'c,
c<d (Fig. 16, I).

EXAMPLES FOR PRACTICE.

1. Construct a shear diagram for the beam represented in
Fig. 10, neglecting the weight of the beam (see example 1, Art. 37).

2. Construct the shear diagram for the beam represented in
Fig. 11, neglecting the weight of the beam (see example 3,
Art 37).

STRENGTH OF MATERIALS

3. Construct the shear diagram for the beam of Fig. 12
when it sustains, in addition to the loads represented, its own
weight, 800 pounds, and a uniform load of 500 pounds per foot
(see example 4, Art. 37).

4. Figs, a, cases 1 and 2, Table B, represent two cantilever
beams, the first bearing a concentrated load P at the free end,
and the second a uniform load W. Figs, b are the corresponding
shear diagrams. Take P and W equal to 1,000 pounds, and satisfy
yourself that the diagrams are correct.

and satisfy yourself that the diagrams are correct.

5. Figs. «, cases 3 and 4, same table, represent simple
beams supported at their ends, the first bearing a concentrated

500 Ibs. looolbs. zooolbs.

£-„ 0 ,

lA IB
i i
j i

!A 'B'

C D*
C1 D'|

Sca-le i"= sooo Ibs. d

Fig. 16.

load P at the middle, and the second a uniform load W. Figs.
b are the corresponding shear diagrams. Take P and W equal
to 1,000 pounds, and satisfy yourself that they are correct.

39. Maximum Shear. It is sometimes desirable to know
the greatest or maximum value of the shear in a given case. This
value can always be found with certainty by constructing the shear
diagram, from which the maximum value of the shear is evident at
a glance. In any case it can most readily be computed if one
knows the section for which the shear is a maximum. The stu-
dent should examine all the shear diagrams in the preceding
articles and those that he has drawn, and see that

1. In cantilevers fixed in a wall, the maximum shear
occurs at the wall.

32 STRENGTH OF MATERIALS

2. In simple beams, the maximum shear occurs at a sec-
tion next to one of the supports.

By the use of these propositions one can determine the value
of the maximum shear without constructing the whole shear
diagram. Thus, it is easily seen (referring to the diagrams, page
53) that for a

Cantilever, end load P, maximum shear=P

« , uniform load W, " " =W

Simple beam, middle load P, " " =±P

" « , uniform « W, « " =±W

40. Bending floment. By bending moment at (or for) a
section of a loaded beam, is meant the algebraic sum of the mo-
ments of all the loads (including weight of beam) and reactions
to the left or right of the section with respect to any point in the
section.

41. Rule of Signs. We follow the rule of signs previously
stated (Art. 29) that the moment of a force which tends to pro-
duce clockwise rotation is plus, and that of a force which tends to
produce counter-clockwise rotation is minus; but in order to get
the same sign for the bending moment whether computed from
the right of left, we change the sign of the sum of the moments
when computed from the loads and reactions on the right. Thus .
for section a, Fig. 8, the algebraic sums of the moments of the
forces are :

when computed from tne left,

-1,000 X 5 + 3,000 X 1—2,000 foot-pounds ;
and when computed from the right,

1,000 X 19-3,000 X 15 + 2,000 X 13 + 2,000 X 1 = + 2,000 foot-
pounds.
The bending moment at section a is -2,000 foot-pounds.

Again, for section o, the algebraic sums of the moments of the
forces are:
when computed from the left,

-1,000 X 22 + 3,000x18-2,000 X 16-2,000 X 4 + 3,000 X 2=

-2,000 foot-pounds;
and when computed from the right,

1,000x2= + 2,000 foot-pounds.
The bending moment at the section is -2.000 foot-pounds.

STRENGTH OF MATERIALS 33

It is usually convenient to compute the bending moment for
a section from the forces to the right or left according as there
are fewer forces (loads and reactions) on the right or left side
of the section.

42. Units. It is customary to express bending moments -in
inch -pounds, but often the foot-pound unit is more convenient.
To reduce foot-pounds to inch-pounds, multiply ~by twelve.

43. Notation. We shall use M to denote bending moment at
any section, and the bending moment at a particular section will
be denoted by that letter subscripted; thus M1? M2, etc., denote
values of the bending moment for sections one, two, etc., feet
from the left end of the beam.

Examples. 1. Compute the bending moments for sections
one foot apart in the beam represented in Fig. 9, neglecting the
weight of the beam. (The right and left reactions are 3,700 and
2,300 pounds respectively. See example 1, Art. 33.)

Since there are no forces acting on the beam to the left of the
left support, M0— 0. To the left of the section one foot from the
left end there is but one force, the left reaction, and its arm is one
foot; hence M, = + 2,300x1=2,300 foot-pounds. To the left of
a section two feet from the left end there are two forces, 2,300 and
1,000 pounds, and their arms are 2 feet and 1 foot respectively;
hence M2= + 2,300x2-1,000x1 =3,600 foot-pounds. At the
left of all sections between B and C there are only twro forces,
2,300 and 1,000 pounds; hence

M3= + 2,300 X 3-1,000 X 2= +4,900 foot-pounds,
M4= + 2,300 X 4-1,000 X 3= + 6,200 «
M5= + 2,300 X 5-1,000 X 4= + 7,500 "
M0= + 2,300x6-1,000x5= + 8,800 ' «

To the right of a section seven feet from the left end there
are two forces, the 3,000-pound load and the right reaction
(3,700 pounds), and their arms with respect to an origin in that
section are respectively one foot and three feet; hence

M7= -(-3,700 X 3 + 3,000 x 1) = + 8,100 foot-pounds.

To the right of any section between E and D there is only one
force, the right reaction ; hence

34 STRENGTH OF MATERIALS

M8=-(-3,700 X 2)^7,400 foot-pounds,
M9 =-(-3,700 X 1)=3,700

Clearly M10=0.

2. A simple beam 10 feet long and supported at its ends
weighs 400 pounds, and bears a uniformly distributed load of 1,600
pounds. Compute the bending moments for sections two feet
apart.

Each reaction equals one-half the whole load, that is, -| of
.(1,600+400) =1,000 pounds, and the load per foot including
weight of the beam is 200 pounds. The forces acting on the
beam to the left of the first section, two feet from the left end, are
the left reaction (1,000 pounds) and the load (including weight)
on the part of the beam to the left of the section (400 pounds).
The arm of the reaction is 2 feet and that of the 400 -pound force
is 1 foot (the distance from the middle of the 400 -pound load to
the section). Hence

M2= + 1,000 X 2-400 X 1 = + 1,600 foot-pounds.

The forces to the left of the next section, 4 feet from the left
end, are the left reaction and all the load (including weight of
beam) to the left (800 pounds). The arm of the reaction is 4 feet,
and that of the 800-pound force is 2 feet; hence

M4= + 1,000 X 4-800 X 2 = + 2,400 foot-pounds.
Without further explanation the student should see that

M6= + 1,000 X 6-1,200 X 3= + 2,400 foot-pounds,
M8= + 1,000 X 8-1,600 X 4= + 1,600

Evidently M0=M10=0.

8. Compute the values oj; the bending moment in example
1, taking into account the weight of the beam, 400 pounds. (The
right and left reactions are respectively 3,900 and 2,500 pounds;
see example 3, Art. 33.)

We proceed as in example 1, except that the moment
of the weight of the beam to the left of each section (or to
the right when computing from forces to the right) must be
included in the respective moment equations. Thus, computing
from the left,

STRENGTH OF MATERIALS

M0 -0

Mt = + 2,500 X 1-40 X j= + 2,480 foot-pounds,

M2 = + 2,500 X 2-1,000 X 1-80 X 1= + 3,920,

M3 = + 2,500 X 3-1,000 X 2-120 X 1£= + 5,320,

M4 = + 2,500 X 4-1,000 X 3-160 X 2=+ 6,680,

M5 = + 2,500 X 5-1,000 X 4-200 X 2±= + 8,000,

M, =+ 2,500 X 6-1,000 X 5-240 X 3= + 9,280.

Computing from the right,

M7 =-(-3,900x3+3,OOOxl + 120x

M8 =-(-3,900x2 + 80xl) = + 7,720,

M9 =-(-3,900xl+40x4)=+3,880,

M10 = 0.

EXAMPLES FOR PRACTICE.

1. Compute the values of the bending moment for sections
one foot apart, beginning one foot from the left end of the
beam represented in Fig. 10, neglecting the weight of the beam.
(The right and left reactions are 3,300 and 4,000 pounds respec-
tively; see example 2, Art. 33.)

M,= - 2,100 M6 = + 3,400 Mn= + 2,100 M16=-6,400
M2= - 4,200 M7 = + 5,300 M12= + 400 M17=-4,800
M3= - 2,300 M. = + 7,200 M13= - 1,300 M18=-3,200

Ans.
(in foot-
pounds)

M4= - 400 M, = + 5,500 M14= - 3,000 M19=-l,600
M5= + 1,500 M10= + 3,800 M15= - 4,700 M20= 0

2. Solve the preceding example, taking into account the
weight of the beam, 42 pounds per foot. (The right and left
reactions are 3,780 and 4,360 pounds respectively; see example 4,
Art. 33.)

'Mx= - 2,121 M6 =+4,084 Mn= + 2,799 M16== - 6,736

Ans. M2= - 4,284 M7 = + 6,071 M12=+ 976 M17= - 4,989

(in foot. J M3= - 2,129 M8 = + 8,016 M13= - 889 M18= - 3,284

pounds) MI== _ 16 M9 = + 6,319 Mu= - 2,796 M19= - 1,621

\ M5= + 2,055 M10=+ 4,580 M15= - 4,745 M20= 0

3. Compute the bending moments for sections one foot
apart, of the beam represented in Fig. 11, neglecting the weight.
(The right and left reactions are 1,444 and 1,556 pounds respect-
ively; see example 1, Art. 33.)

STRENGTH OF MATERIALS

pounds)

,= + l,556 M5= + 5,980 M9 =+6,104 M,,= +4,328

Ma=+<M>68 M7=+6,992 Mn= + 5,216 MI6= + 1,440
M4= + 5,024 M8= + 6,548 M12= + 4,772 M16= 0

4 Compute the bending moments at sections one foot apart
in the beam of Fig. 12, taking into account the weight of the beam,
800 pounds, and a uniform load of 500 pounds per foot. (The
right and left reactions are 4,870 and 11,930 pounds respectively;
see Exs. 3 and 4, Art. 33.)

[X^- 270 M6 =-19,720 Mn=+ 3,980 M16=12,180

Ans. Ma=- 3,080 M7 = -13,300 M12=+ 6,700 M17= 12,200

(in foot J M8=- 6,430 M8 =- 7,420 M13=+ 8,880 M18= 8,680

pounds) M4= -10,320 M9 =- 3,080 M14= + 10,520 M19= 4,620

L M6= -14,750 Mlo= + 720 M1B= + 11,620 M20= 0

44. Moment Diagrams. The way in which the bending
moment varies from section to section in a loaded beam can be
well represented by means of a diagram called a moment diagram.
To construct such a diagram for any loaded beam,

1000 Ibs.

sooolbs. 3000 Itas.
2'

ScaJie:i « looooft.-lbs
Fig. 17.

1. Lay off a base-line just as for a shear diagram (see
Art. 38).

2. Draw a line such that the distance from any point of it
to the base-line equals (by some scale) the value of the bending
moment at the corresponding section of the beam, and so that the
line is above the base where the bending moment is positive and
below it where it is negative. (This line is called a "moment
line."}

STRENGTH OF MATERIALS 37

Examples. 1. It is required to construct a moment dia-
gram for the beam of Fig. 17, a (a copy of Fig. 9), loaded as
there shown.

Layoff A'E' (Fig. 17, 5) as a base. In example 1, Art. 43,
we computed the values of the bending moment for sections one
foot apart, so we erect ordinates at points of A'E' one foot apart,
to represent the bending moments.

We shall use a scale of 10,000 foot-pounds to the inch; then
<he ordinates (see example 1, Art. 43, for values of M) will be:
One foot from left end, 2,300-^-10,000 = 0.23 inch,
Two feet « « " 3,600-^-10,000 = 0.36 "
Three " « « « 4,900-^10,000 = 0.49 "
Four " « « " 6,200-^-10,000 = 0.62 "
etc., etc.

•r

looolbs.
I 5'

2000 Ibs.

30OO Ib3.
«i 1 -i J

| 3

- T *

i
1

.^TTTTrfflTn'niTnii

TTfTTTTTTTTnUi

B1 C1 D1 E'

ScoJe: i"- looooft.-lbs.
Fig. 18.

Laying these ordinates off, and joining their ends in succession,
we get the line A'&<%£E', which is the bending moment line.
Fig. 17, b, is the moment diagram.

2. It is required to construct the moment diagram for the
beam, Fig. 18, a (a copy of Fig. 9), taking into account the weight
of -the beam, 400 pounds.

The values of the bending moment for sections one foot apart
were computed in example 3, Art. 43. So we have only to lay off
ordinates equal to those values, one foot apart, on the base A'E;
(Fig. 18, b).

To a scale of 10,000 foot-pounds to the inch the ordinates
(see example 3, Art. 43, for values of M) are:

STRENGTH OF MATERIALS

At left end, 0

One foot from left end, 2,480-^10,000=0.248 inch
Two feet « " « 3,920-^-10,000=0.392 «
Three « " « « 5,320-f- 10,000=0.532 «
Four " « « « 6,680-^-10,000=0.668 «
Laying these ordinates off at the proper points, we get
as the moment line.

3. It is required to construct the moment diagram for the
cantilever beam represented in Fig. 19, #, neglecting the weight
of the beam. The bending moment at B equals

-500x2=-l,000 foot-pounds;
atC,

-500 X 5-1,000 X 3=-5,500 ;
and at D,

-500 X 9-l,OOOX 7-2,000 X 4=-19,500.

500 Ibs. looolbs.

Sca,1e:i"* sooooft.-ltas

Fig. 19.

Using a sca/e of 20,000 foot-pounds to one inch, the ordinates
in the bending moment diagram are:

AtB, 1,000-5-20,000=0.05 inch,

« C, 5,500-^20,000=0.275 «

« D, 19,500^-20,000=0.975 «

Hence we lay these ordinates off, and downward because the bend-
ing moments are negative, thus fixing the points 5, c and d. The
bending moment at A is zero; hence the moment line connects A
&, c and d. Further, the portions A£, be and cd are straight, as
can be shown by computing values of the bending moment for
sections in AB, BC and CD, and laying off the corresponding
ordinates in the moment diagram.

STRENGTH OF MATERIALS

4. Suppose that the cantilever of the preceding illustration
sustains also a uniform load of 100 pounds per foot (see Fig. 20, a).
Construct a moment diagram.

First, we compute the values of the bending moment at sev-
eral sections; thus,

Mx=-500 X 1-100 X J=-550 foot-pounda,
Ma=-500x 2-200 XI =-1,200,
Ms=-500 X 3-1,000 x 1-300 X 1 J=-2,950,
M4=-500 X 4-1,000 X 2-400 X 2=-4,800,
M5=-500 X 5-1,000 X 3-500 X 2J= -6,750,
M6=-500 X 6-1,000 X 4-2,000 X 1-600 X 3=-10,800,
M°=:-500 X 7-1,000 X 5-2,000 X 2-700 X 3£=-14,950,
M8=-500 X 8-1,000 X 6-2,000 X 3-800 X 4=-19,200,
1L=-500 X 9-1,000 X 7-2,000 X 4-900 X 4£=-23,550.

500 Ibs. 1000 Ibs.

2000 Ibs.

ScaJe:

Fig. 20.

These values all being negative, the ordinates are all laid off
downwards. To a scale of 20,000 foot-pounds to one inch, they
fix the moment line A.'bcd.

EXAflPLES FOR PRACTICE.

1. Construct a moment diagram for the beam represented in
Fig. 10, neglecting the weight of the beam. (See example 1,
\rt. 43).

2. Construct a moment diagram for the beam represented
in Fig. 11, neglecting the weight of the beam. (See example 3,
Art. 43).

3. Construct the moment diagram for the beam of Fig. 12

40 STRENGTH OF MATERIALS

when it sustains, in addition to the loads represented and its own
weight (800 pounds), a uniform load of 500 pounds per foot.
(See example 4, Art. 43.)

4. Figs, a, cases 1 and 2, page 53, represent two cantilever
beams, the first bearing a load P at the free end, and the second
a uniform load W. Figs, c are the corresponding moment
diagrams. Take P and "W equal to 1,000 pounds, and I equal to
10 feet, and satisfy yourself that the diagrams are correct.

5. Figs. #, cases 3 and 4, page 53, represent simple beams
on end supports, the first bearing a middle load P, and the other a
uniform load W. Figs, c are the corresponding moment dia-
grams. Take P and "W equal to 1,000 pounds, and I equal to
10 feet, and satisfy yourself that the diagrams are correct.

45. Maximum Bending Moment. It is sometimes desirable
to know the greatest or maximum value of the bending moment
in a given case. This value can always be found with certainty
by constructing the moment diagram, from which the maximum
value of the bending moment is evident at a glance. But in any
case, it can be most readily computed if one knows the section for
which the bending moment is greatest. If the student will com-
pare the corresponding shear and moment diagrams which have
been constructed in foregoing articles (Figs. 13 and 17, 14 and
18, 15 and 19, 16 and 20), and those which he has drawn, he will
see that — The maximum bending moment in a beam occurs
where the shear changes sign.

By the help of the foregoing principle we can readily com-
pute the maximum moment in a given case. We have only to
construct the shear line, and observe from it where the shear
changes sign; then compute the bending moment for that section.
If a simple beam has one or more overhanging ends, then the shear
changes sign more than once — twice if there is one overhanging
end, and three times if two. In such cases we compute the
bending moment for each section where the shear changes sign;
the largest of the values of these bending moments is the maxi-
mum for the beam.

The section of maximum bending moment in a cantilever
fixed at one end (as when built int3 a wall) is always at the wall.

STRENGTH OF MATERIALS 41

Thus, without reference to the moment diagrams, it is readily seen
that,

for a cantilever whose length is Z,

with an end load P, the maximum moment is PZ,
" a uniform « W, " " " " J "W7.

Also by the principle, it is seen that,

for a beam whose length is Z, on end supports,

with a middle load P, the maximum moment is ^ PZ,
« uniform " W, « " " " J- WZ.

46. Table of Maximum Shears, Moments, etc. Table B
on page 53 shows the shear and moment diagrams for eight
simple cases of beams. The first two cases are built-in cantilevers;
the next four, simple beams on end supports; and the last two,
restrained beams built in walls at each end. In each case I
denotes the length.

CENTER OF GRAVITY AND HOMENT OF INERTIA.

It will be shown later that the strength of a beam depends
partly on the form of its cross-section. The following discussion
relates principally to cross -sections of beams, and the results
reached (like shear and bending moment) will be made use of
later in the subject of strength of beams.

47. Center of Gravity of an Area. The student probably
knows what is meant by, and how to find, the center of gravity of
any flat disk, as a piece of tin. Probably his way is to balance
the piece of tin on a pencil point, the point of the tin at which it so
balances being the center of gravity. (Really it is midway between
the surfaces of the tin and over the balancing point.) The center
of gravity of the piece of tin, is also that point of it through which
the resultant force of gravity on the tin (that is, the weight of the
piece) acts.

By "center of gravity" of a plane area of any shape we mean
that point of it which corresponds to the center of gravity of a
piece of tin when the latter is cut out in the shape of the area.
The center of gravity of a quite irregular p,rea can be found most
readily by balancing a piece of tin or stiff paper cut in the shape
of the area. But when an area is simple in shape, or consists of
parts which are simple, the center of gravity of the whole can be

42 STRENGTH OF MATERIALS

found readily by computation, and such a method will now be
described.

48. Principle of floments Applied to Areas. Let Fig. 21
represent a piece of tin which has been divided off into any num-
ber of parts in any way, the weight of the whole being "W, and
that of the parts "W,, W2, W3, etc. Let Ow C2, C8, etc., be the
centers of gravity of the parts, C that of the whole, and <?n <?2, <?3,
etc., and c the distances from those centers of gravity respectively
to some line (L L) in the plane
of the sheet of tin. When the
tin is lying in a horizontal posi-
tion, the moment of the weight
of the entire piece about L L is
"We, and the moments of the
parts are W^, W2<?2, etc. Since
the weight of the whole is the .

resultant of the weights of the

o Fig. 21.

parts, the moment of the weight

of the whole equals the sum of the moments of the weights of the
parts; that is,

Now let A:, A2, etc. denote the areas of the parts of the pieces
of tin, and A the area of the whole; then since the weights are
proportional to the areas, we can replace the "W's in the preceding
equation by corresponding A's, thus:

Ac=A1<?1 + A2<?2-fetc ..... (4)

If we call the product of an area and the distance of its
center of gravity from some line in its plane, the "moment" of the
area with respect to that line, then the preceding equation may be
stated in words thus:

The moment of an area with respect to any line equals the
algebraic sum of the moments of the parts of the area.

If all the centers of gravity are on one side of the line with
respect to which moments are taken, then all the moments should be
given the plus sign; but if some centers of gravity are on one side
and some on the other side of the line, then the moments of the
areas whose centers of gravity are on one side should be given the

STRENGTH OF MATERIALS 43

same sign, and the moments of the others the opposite sign. The
foregoing is the principle of moments for areas, and it is the basis
of all rules for finding the center of gravity of an area.

To find the center of gravity of an area which can be divided
np into simple parts, we write the principle in forms of equations
for two different lines as "axes of moments," and then solve the
equations for the unknown distances of the center of gravity of the
whole from the two lines. We explain further by means of specific
examples.

Examples. 1. It is required to find the center of gravity
of Fig. 22, a, the width being uniformly one inch.

The area can be divided into two rectangles. Let Ct and

•C

•C2

^ To

Fig. 22.

C2 be the centers of gravity of two such parts, and C the center of
gravity of the whole. Also let a and ~b denote the distances of C
from the two lines OL' and OL" respectively.

The areas of the parts are 6 and 3 square inches, and their
arms with respect to OL' are 4 inches and \ inch respectively, and
with respect to OL" -J inch and 1^ inches. Hence the equations of
moments with respect to OL' and OL" (the whole area being 9
square inches) are:

9x^=6x1+3x14=75.

Hence, a — 25.5-^-9 — 2.83 inches,

I = 7.5-V-9 = 0.83 « .

2. It is required to locate the center of gravity of Fig. 22, b,
the width being uniformly one inch.

STRENGTH OF MATERIALS

The figure can be divided up into three rectangles. Let Cj, C2
and C3 be the centers of gravity of such parts, C the center of
gravity of the whole; and let a denote the (unknown) distance of
C from the base. The areas of the parts are 4, 3 0 and 4 square
inches, and their " arms " with respect to
the base are 2,^ and 2 inches respectively;
hence the equation of moments with re-
spect to the base (the en tire area being 18
square inches) is:

18X0 = 4x2+10X-J+4x2 = 21.
Hence, a — 21-r-18 — 1.17 inches.

From the symmetry of the area it is plain
that the center of gravity is midway be-
tween the sides.

EXAMPLE FOR PRACTICE.

1. Locate the center of gravity of
Fig. 23. Fig. 23.

Ans. 2.3 inches above the base.

49. Center of Gravity of Built=up Sections. In Fig. 24
there are represented cross-sections of various kinds of rolled steel,
called "shape steel," which is used extensively in steel construction.
Manufacturers of this material publish "handbooks" giving full
information in regard thereto, among other things, the position of
the center of gravity of each cross section. With such a handbook

1-foea.m

Channel

Angle

Z-b^r

Fig. 2L

available, it is therefore not necessary actually to compute the posi-
tion of the center of gravity of any section, as we did in the pre-
ceding article; but sometimes several shapes are riveted together to

STRENGTH OF MATERIALS

make a "built-up" section (see Fig. 25), and then it may be neces-
sary to compute the position of the center of gravity of the section.
Example. It is desired to locate the center of gravity of the
section of a built-up beam represented in Fig. 25. The beam con-

14"-

CVJ

Fig. 25.

sists of two channels and a plate, the area of the cross -section of a

channel being 6.03 square inches.

Evidently the center of gravity of each channel section is 6

inches, and that of the plate section is 12^ inches, from the bottom.

Let c denote the dis-
tance of the center of
gravity of the whole
section from the bot-
tom; then since the
area of the plate section
is 7 square inches, and
that of the whole sec-
tion is 19.06,

19.06X^ = 6.03X6+
6.03X6 + 7 X12J =
158.11.
Hence, c= 158.11-^-19.06=8.30 inches, (about).

EXAMPLES FOR PRACTICE.

1. Locate the center of gravity of the built-up section of

26-

46 STRENGTH OF MATERIALS

Fig. 26, #, the area of each "angle" being 5.06 square inches, and
the center of gravity of each being as shown in Fig. 26, b.

Ans. Distance from top, 3.08 inches.

2. Omit the left-hand angle in Fig. 26, 0, and locate the
Center of gravity of the remainder.

. ( Distance from top, 3.65 inches,
'" | " " left side, 1.19 inches.

50. floment of Inertia. If a plane area be divided into an
infinite number of infinitesimal parts, then the sum of the prod-
ucts obtained by multiplying the area of each part by the square
of its distance from some line is ealled the moment of inertia, of the
area with respect to the line. The line to which the distances are
measured is called the inertia-axis; it may be taken anywhere in
the plane of the area. In the subject of beams (where we have
sometimes to compute the moment of inertia of the cross -section
of a beam), the inertia-axis is taken through the center of gravity
of the section and horizontal.

An approximate value of the moment of inertia of an area
can be obtained by dividing the area into small parts (not infini-
tesimal), and adding the products obtained by multiplying the
area of each part by the square of the distance from its center to
the inertia-axis.

Example. If the rectangle of Fig. 27, #, is divided into 8
parts as shown, the area of each is one square inch, and the dis-
tances from the axis to the centers of gravity of the parts are \
and 1^ inches. For the four parts lying nearest the axis the
product (area times distance squared) is:

lX( -J)2=4; and for the other parts it is

Hence the approximate value of the moment of inertia of the area
with respect to the axis, is

If the area is divided into 32 parts, as shown in Fig. 27, J,
the area of each part is J square inch. For the eight of the little
squares farthest away from the axis, the distance from their centers
of gravity to the axis is 1J inches; for the next eight it is 1J;
for the next eight J; and for the remainder J inch. Hence an

STRENGTH OF MATERIALS

approximate value of the moment of inertia of the rectangle with
respect to the axis is :

,2!L

If we divide the rectangle into still smaller parts and form
the products

(small area) X (distance)2,

and add the products just as we have done, we shall get a larger
answer than ,10J. The smaller the parts into which the rec-
tangle is divided, the larger will be the answer, but it will never
be larger than 10§. This 10§ is the sum corresponding to a

division of the rectangle into an
infinitely large number of parts
(infinitely small) and it is the
exact value of the moment of
inertia of the rectangle with re-
spect to the axis selected.

There are short methods of
computing the exact values of the
„. 9 moments of inertia of simple fig-

ures (rectangles, circles, etc.,),

but they cannot be given here since they involve the use of difficult
mathematics. The foregoing method to obtain approximate val-
ues of moments of inertia is used especially when the area is quite
irregular in shape, but it is given here to explain to the student
the meaning of the moment of inertia of an area. He should
understand now that the moment of inertia of an area is sim-
ply a name for such sums as we have just computed. The name
is not a fitting one, since the sum has nothing whatever to do with
inertia. It was first used in this connection because the sum is
very similar to certain other sums which had previously been
called moments of inertia.

51. Unit of Moment of Inertia. The product (area X dis-
tance2) is really the product of four lengths, two in each factor ;
and since a moment of inertia is the sum of such products, a
moment of inertia is also the product of four lengths. Now the
product of two lengths is an area, the product of three is a vol-
ume, and the product of four is moment of inertia — unthinkable in

48 STRENGTH OF MATERIALS

the way in which we can think of an area or volume, and there-
fore the source of much difficulty to the student. The units of
these quantities (area, volume, and moment of inertia) are respec-
tively :

the square inch, square foot, etc.,
" cubic " , cubic " " ,
" biquadratic inch, biquadratic foot, etc.;

but the biquadratic inch is almost exclusively used in this connec-
tion; that is, the inch is used to compute
values of moments of inertia, as in the pre-
ceding illustration. It is often written
thus: Inches4,
•i 52. Moment of Inertia of a Rectangle.

w -

Let 1} denote the base of a rectangle, and a
Pig 28. its altitude; then by higher mathematics it

can be shown that the moment of inertia

of the rectangle with respect to a line through its center of gravity
and parallel to its base, is y1^ W.

Example. Compute the value of the moment of inertia of
a rectangle 4x12 inches with respect to a line through its center
of gravity and parallel to the long side.

Here 5=12, and a = 4 inches ; hence the moment of inertia
desired equals

TV(12X43)= 64 inches4.

EXAHPLE FOR PRACTICE.

1. Compute the moment of inertia of a rectangle 4x12
inches with respect to a line through its center of gravity and
parallel to the short side. Ans. 576 inches4.

53. Reduction Formula. In the previously mentioned
"handbooks" there can be found tables of moments of inertia of
all the cross-sections of the kinds and sizes of rolled shapes made.
The inertia-axes in those tables are always taken through the cen-
ter of gravity of the section, and usually parallel to some edge of
the section. Sometimes it is necessary to compute the moment of
inertia of a "rolled section" with respect to some other axis, and
if the two axes (that is, the one given in the tables, and the other)
are parallel, then the desired moment of inertia can be easily com-
puted from the one given in the tables by the following rule:

STRENGTH OF MATERIALS

-co

The moment of inertia of an area with respect to any axis
equals the moment of inertia with respect to a parallel axis
through the center of gravity r, plus the product of the area and
the square of the distance between the axes.

Or. if I denotes the moment of inertia with respect to any axis;
I0 the moment of inertia with respect to a parallel axis through
the center of gravity; A the area; and d the (^.stance between the
axes, then

I=Io+A^.... (5)

Example. It is required to compute the moment of inertia
of a rectangle 2x8 inches with respect to a line parallel to the
long side and 4 inches from the center of gravity.

Let I denote the moment of inertia sought, and I0 the moment
of inertia of the rectangle with respect
to a line parallel to the long side and
through the center of gravity (see Fig.
28). Then

I0=^5a8 (see Art. 52); and,
since £=8 inches and a=2 inches,
Io=Tiir(8x23)=51L biquadratic inches.

The distance between the two inertia-
axes is 4 inches, and the area of the
rectangle is 16 square inches, hence
equation 5 becomes Fi£' 29»

I=5iL-f 16x42=261iL biquadratic inches.

EXAMPLE FOR PRACTICE,

1. The moment of inertia of an "angle" 2-|x2X-J inches
(lengths of sides and width respectively) with respect to a line
through the center of gravity and parallel to the long side, is 0.64
inches*. The area of the section is 2 square inches, and the dis-
tance from the center of gravity to the long side is 0.63 inches.
(These values are taken from a "handbook".) It is required to
compute the moment of inertia of the section with respect to a
line parallel to the long side and 4 inches from the center of
gravity. Ans. 32.64 inches*.

54. Moment of Inertia of Built-up Sections. As before
stated, beams are sometimes "built up" of rolled shapes (angles,

STRENGTH OF MATERIALS

channels, etc.). The moment of inertia of such a section with
respect to a definite axis is computed by adding the moments of
inertia of the parts, all with respect to that same axis. This is the
method for computing the moment of any area which can be
divided into simple parts.

The moment of inertia of an area which may be regarded as
consisting of a larger area minus other areas, is computed by sub-
tracting from the moment of inertia of the large area those of the
"minus areas."

Examples. 1. Compute the moment of inertia of the built-
up section represented in Fig. 30 (in part same as Fig. 25) with
respect to a horizontal axis
passing through the center
of gravity, it being given
that the moment of inertia
of each channel section
with respect to a horizontal *§"
axis through its center of
gravity is 128.1 inches4,
and its area 6.03 square
inches.

The center of gravity of
the whole section was found

in the example of Art. 49 to be 8.30 inches from the bottom of
the section; hence the distances from the inertia-axis to the
centers of gravity of the channel section and the plate are 2.30
and 3.95 inches respectively (see Fig. 30).

The moment of inertia of one channel section with respect to
the axis A A (see equation 5, Art. 53) is:

128.1 + 6.03 X2.302=160.00 inches4.

The moment of inertia of the plate section (rectangle) with re-
spect to the line a" a" (see Art. 52) is:

^ ba3=rV[14>< (i)3H°-15 inches4;
and with respect to the axis A A (the area being 7 square inches)

it is:

0.15+ 7 X3.952=109.37 inches4.

Therefore the moment of inertia of the whole section with re-

spect to AA is :

2x160.00+109.37=429.37 inches*.

STRENGTH OF MATERIALS

2. It is required to compute the moment of inertia of the
" hollow rectangle " of Fig. 29 with respect to a line through the
center of gravity and parallel to the short side.

The amount of inertia of the large rectangle with respect to
the named axis (see Art. 52) is:

<pi.66" a.

Fig. 31.

and the moment of inertia of the smaller one with respect to the
same axis is:

TV(4X8°) = 170§;

hence the moment of inertia of the hollow section with respect
to the axis is:

416§ - 170§ =-246 inches4.
EXAMPLES FOR PRACTICE.

1. Compute the moment of inertia of the section repre-
sented in Fig. 31, <z, about the axis AA, it being 3.08 inches
from the top. Given also that the area of one angle section is
5.06 square inches, its center of gravity C (Fig. 31, b) 1.66 inches
from the top, and its moment of inertia with respect to the axis aa
17.68 inches4. Ans. 145.8 inches*.

2, Compute the moment of inertia of the section of Fig. 31, a,

STRENGTH OF MATERIALS

with respect to the axis BB. Given that distance of the center
of gravity of one angle from one side is 1.66 inches (see Fig. 31,5),
and its moment of inertia with respect to bb 17.68 inches.

Ans. 77.618 inches4.

55. Table of Centers of Gravity and floments of Inertia.
Column 2 in Table A below gives the formula for moment of
inertia with respect to the horizontal line through the center of
gravity. The numbers in the third column are explained in Art.
62; and those in the fourth, in Art. 80.

TABLE A.

Moments of Inertia, Section Moduli, and Radii of Gyration.

In each case the axis is horizontal and passes through the center of gravity.

Section.

Moment of
Inertia.

Section
Modulus.

Radius of
Gyration.

a4
12

1/12

ba3
12

0.049d*

0.049 (d4-d/)

baf
6

0.098d3

0.098-

l/d24-d,2

STRENGTH OF BEAMS.

56. Kinds of Loads Considered. The loads that are applied
to a horizontal beam are usually vertical, but sometimes forces are
applied otherwise than at right angles to beams. Forces acting on
beams at right angles are called transverse forces ; those applied

STRENGTH OF MATERIALS

TABLE B.

Shear Diagrams (b) and Moment Diagrams (c) for Eight Different Cases (a).
Also Values of Maximum Shear (V), Bending floment (M), and Deflection (d).

'. f

V=P, M=P1, d=P!3H-3EI.

d=W!3-H8EI.

JW-anfform

, M=MP1, d= P18-S-48EI.

=J^ W, M=%W1, d=5Wl3-j-384EI.

, M=Pab-l.

V=P, M=Pa, d=Pa(312-4a2)-?-24EI.

d=P!3-f-192EI.

STRENGTH OF MATERIALS

parallel to a beam are .called longitudinal forces ; and others are
called inclined forces. For the present we deal only with beams
subjected to transverse forces (loads and reactions).

57. Neutral Surface, Neutral Line, and Neutral Axis. When
a beam is loaded it may be wholly convex up (concave down), as a
cantilever; wholly convex down (concave up), as a simple beam
on end supports; or partly convex up and partly convex down, as
a simple beam with overhanging ends, a restrained beam, or a con-

Fig. 32.

tinuous beam. Two vertical parallel lines drawn close together on
the side of a beam before it is loaded will not be parallel after it
is loaded and bent. If they are on a convex-down portion of a
beam, they will be closer at the top and farther apart below than
when drawrn (Fig. 32&), and if they are on a convex-up portion,
they will be closer below and farther apart above than when drawn
(Fig. 325).

The " fibres " on the convex side of a beam are stretched and
therefore under tension, while those on the concave side are short-
ened and therefore under compression. Obviously there must be
some intermediate fibres which are neither stretched nor shortened,
i. e., under neither tension nor compression. These make up
a sheet of fibres and define a surface in the beam, which surface is
called the neutral surface of the beam. The intersection of the
neutral surface with either side of the beam is called the neutral
line, and its intersection with any cross-section of the beam is
called the neutral axis of that section. Thus, if ab is a fibre that
has been neither lengthened nor shortened with the bending of the
beam, then nn is a portion of the neutral line of the beam; and,
if Fig. 32<? be taken to represent a cross-section of the beam, NN
is the neutral axis of the section.

It can be proved that the neutral axis of any cross-section of

STRENGTH OF MATERIALS

a loaded bearn passes through the center of gravity of that section,
provided that all the forces applied to the beam are transverse, and
that the tensile and compressive stresses at the cross-section are
all within the elastic limit of the material of the beam.

58. Kinds- of Stress at a Cross=section of a Beam. It has
already been explained in the preceding article that there are ten-
sile and compressive stresses in a beam, and that the tensions are
on the convex side of the beam and the compressions on the con-
cave (see Fig. 33). The forces T and C are exerted upon the
portion of the beam represented by the adjoining portion to the

i T

Fig. 33.

right (not shown). These, the student is reminded, are often called
fibre stresses.

Besides the fibre stresses there is, in general, a shearing stress
at every cross-section of a beam. This may be proved as follows:

Fig. 34 represents a simple beam on end supports which has
actually been cut into two parts as shown. The twro parts can
maintain loads when in a horizontal position, if forces are applied
at the cut ends equivalent to the forces that would act there if the
beam were not cut. Evidently in the solid beam there are at the
section a compression above and a tension below, and such forces
can be applied in the cut beam by means of a short block C and a
chain or cord T, as shown. The block furnishes the compressive
forces and the chain the tensile forces. At first sight it appears as
if the beam would stand up under its load after the block and
chain have been put into place. Except in certain cases*, how-
ever, it would not remain in a horizontal position, as would the

* When the external shear for the section is zero.

STRENGTH OF MATERIALS

solid beam. This shows that the forces exerted by the block and
chain (horizontal compression and tension ) are not equivalent to
the actual stresses in the solid beam. What is needed is a vertical
force at each cut end.

Suppose that Rj is less than L1 + L2 + weight of A, i. e., that
the external shear for the section is negative; then, if vertical' pulls
be applied at the cut ends, upward on A and downward on B, the
beam will stand under its load and in a horizontal position, just as
a solid beam. These pulls can be supplied, in the model of the
beam, by means of a cord S tied to two brackets fastened on A and

Fig. 34.

///ft/// RI

Fig. 35.

B, as shown. In the solid beam the two parts act upon each
other directly, and the vertical forces are shearing stresses, since
they act in the plane of the surfaces to which they are applied.

59. Relation Between the Stress at a Section and the Loads
and Reactions on Either Side of It. Let Fig. 35 represent the
portion of a beam on the left of a section ; and let Rx denote the
left reaction; Lj and L2 the loads; W the weight of the left part;

C, T, and S the compression, tension, and shear respectively which
the right part exerts upon the left.

Since the part of the beam here represented is at rest, all the
forces exerted upon it are balanced; and when a number of hori-
zontal and vertical forces are balanced, then

1. The algebraic sum of the horizontal forces equals zero.

2. " « " " " vertical " " "

3. " " " " " moments of all the forces with respect to
any point equals zero.

To satisfy condition 1, since the tension and compression are

the only horizontal forces, the tension must equal the compression.

To satisfy condition 2, S (the internal shear) must equal the

STRENGTH OF MATERIALS 57

algebraic sum of all the other vertical forces on the portion, that
is, must equal the external shear for the section; also, S must act
up or down according as the external shear is negative or positive.
In other words, briefly expressed, the internal and external shears
at a section are equal and opposite*

To satisfy condition 3, the algebraic sum of the moments of
the fibre stresses about the neutral axis must be equal to the sum
of the moments of all the other forces acting on the portion about
the same line, and the signs of those sums must be opposite. (The
moment of the shear about the neutral axis is zero.) Now, the
sum of the moments of the loads and reactions is called the bend-
ing moment at the section, and if we use the term resisting mo-
ment to signify the sum of the moments of the fibre stresses (ten-
sions and compressions ) about the neutral axis, then we may say
briefly that the resisting and the bending moments at a section are
equal, and the two moments are opposite in sign.

60. The Fibre Stress. As before stated, the fibre stress is
not a uniform one, that is, it is not uniformly distributed over the
section on which it acts. At any section, the compression is most
" intense " (or the unit-compressive stress is greatest) on the con-
cave side; the tension is most intense (or the unit-tensile stress is
greatest) on the convex side; and the unit-compressive and unit-
tensile stresses decrease toward the neutral axis, at which place the
unit-fibre stress is zero.

If the fibre stresses are within the elastic limit, then the two
straight lines on the side of a beam referred to in Art. 57 will still
be straight after the beam is bent; hence the elongations and short-
enings of the fibres vary directly as their distance from the neutral
axis. Since the stresses (if within the elastic limit) and deforma-
tions in a given material are proportional, the unit-fibre stress
varies as the distance from the neutral axis.

Let Fig. 36& represent a portion of a bent beam, 365 its cross-
section, nn the neutral line, and NN the neutral axis. The way
in which the unit-fibre stress on the section varies can be rep-
resented graphically as follows: Lay off ac, by some scale, to
represent the unit-fibre stress in the top fibre, and join c and n,
extending the line to the lower side of the beam ; also make be* equal
to be? and draw ncf. Then the arrows represent the unit-fibre
stresses, for their ^ngthsvary as their distances from the neutral axis.

STRENGTH OF MATERIALS

6i. Value of the Resisting Moment. If S denotes the unit-
fibre stress in the fibre farthest from the neutral axis (the greatest
unit-fibre stress on the cross-section), and c the distance from the
neutral axis to the remotest fibre, while Sn S2, S3, etc., denote the
unk-fibre stresses at points whose distances from the neutral axis
are, respectively, y,, y^ y^ etc. (see Fig. 36 £), then

S : S, :: c : y,; or S, = -~yr

Q QJ

Also, S2 = ~ y2 ; S3 =—yu etc.

Let

etc., be the areas of the cross-sections of the fibres
S c

dw 10

Fig. 36.

whose distances from the neutral axis are, respectively, yl9 y^ y^
etc. Then the stresses on those fibres are, respectively,
Sj al9 S2 ^2, S3 «3, etc.;

S S S

Or, y\&V ^2^2J — ^3^3? etC*

c c c

The arms of these forces or stresses with respect to the neutrai axis
are, respectively, y^ y^ y^ etc.; hence their moments are

S „ S „ S

etc.,

and the sum of the moments (that is, the resisting moment) is

Q Ql Q

— «i y! +-7^2 y\ + etc- = v^1 y* ~4~a*$ + etc>)

Now ^ y\ + a2 y\ -f etc. is the sum of the products obtained by
multiplying each infinitesimal part of the area of the cross-section
by the square of its distance from the neutral axis; hence, it is the
moment of inertia of the cross-section with respect to the neutral
axis. If this moment is denoted by I, then the value of the resist-

f. SI
ing moment is — o

STRENGTH OF MATERIALS,

PART II.

STRENGTH OF BEANS— (Concluded).

62. First Beam Formula. As shown in the preceding
article, the resisting and bending moments for any section of a
beam are equal; hence

? = M, (6)

all the symbols referring to the same section. This is the most
important formula relating to beams, and will be called the " first
beam formula."

The ratio I H- c is now quite generally called the section
modulus. Observe that for a given beam it depends only on the
dimensions of the cross-section, and not on the material or any-
thing else. Since I is the product of four lengths (see article 51),
I -r- c is the product of three; and hence a section modulus can be
expressed in units of volume. The cubic inch is practically always
used; and in this connection it is written thus, inches3. See Table
A, page 52, for values of the section moduli of a few simple sections.

63. Applications of the First Beam Formula. There are
three principal applications of equation 6, which will now be ex-
plained and illustrated.

64. First Application. The dimensions of a beam and its
manner of loading and support are given, and it is required to
compute the greatest unit-tensile and compressive stresses in the
beam.

This problem can be solved by means of equation 6, written
in this form,

Q MC M

or~

Unless otherwise stated, we assume that the beams are uniform
in cross-section, as they usually are; then the section modulus
(I-s-c) is the same for all sections, and S (the unit-fibre stress on

60 STRENGTH OF MATERIALS

the remotest fibre) varies just as M varies, and is therefore greatest
where M is a maximum.* Hence, to compute the value of the
greatest unit-fibre stress in a given case, substitute the values of
the section modulus and the maximum bending moment in the
preceding equation, and reduce.

If the neutral axis is equally distant from the highest and low-
est fibres, then the greatest tensile and compressive unit- stresses
are equal, and their value is S. If the neutral axis is unequally
distant from the highest and lowest fibres, let c denote its distance
from the nearer of the two, and S' the unit-fibre stress there.
Then, since the unit -stresses in a cross -section are proportional to
the distances from the neutral axis,

If the remotest fibre is on the convex side of the beam, S is tensile
and S' compressive; if the remotest fibre is on the concave side, S
is compressive and S' tensile.

Examples. 1. A beam 10 feet long is supported at its ends,
and sustains a load of 4,000 pounds two feet from the left end
(Fig. 37, &). If the beam is 4 X 12 inches in cross-section (the
long side vertical as usual), compute the maximum tensile and
compressive unit-stresses.

The section modulus of a rectangle whose base and altitude
are 5 and a respectively (see Table A, page 52), is -J&&2; hence,
for the beam under consideration, the modulus is

i- X 4 X 122 = 96 inches3.

To compute the maximum bending moment, we have, first, to find
the dangerous section. This section is where the shear changes
sign (see article 45); hence, we have to construct the shear dia-
gram, or as much thereof as is needed to find where the change of
sign occurs. Therefore we need the values of the reaction.
Neglecting the weight of the beam, the moment equation with
origin at C (Fig. 37, a) is

R, X 10 - 4,000 X 8 = 0, or Rt = 3,200 pounds

* NOTE. Because S is greatest in the section where M is maximum, this
section is usually called the " dangerous section " of the beam.

STRENGTH OF MATERIALS

Then, constructing the shear diagram, we see (Fig. 37, 5) that the

change of sign of the shear (also the dangerous section) is at the

load. The value of the bending moment there is

3,200 X 2 = 6,400 foot-pounds,

6,400 X 12 = 76,800 inch-pounds.

Substituting in equation 6', we find that
76,800

96
4oooltas.

= 800

inch.

U-a'-

c(a)

c'(b)

V /

Fig. 37.

20 It is desired to take into account the weight of the beam
in the preceding example, supposing the beam to be wooden.
The volume of the beam is

— X 10 = 3 J cubic feet ;

and supposing the timber to weigh 45 pounds per cubic foot, the
beam weighs 150 pounds (insignificant compared to the load).
The left reaction, therefore, is

3,200 +(-i-X 150) = 3,275;

62 STRENGTH OF MATERIALS

and the shear diagram looks like Fig. 37, c, the shear changing
sign at the load as before. The weight of the beam to the left of
the dangerous section is 80 pounds; hence the maximum bending
moment equals

3,275 X 2 - 30 X 1 = 6,520 foot-pounds,
or 6,520 X 12 = 78,240 inch-pounds.

Substituting in equation 6', we find that

78,240
S = — q^ — — 815 pounds per square inch.

The weight of the beam therefore increases the unit -stress pro-
duced by the load at the dangerous section by 15 pounds per
square inch.

3. A T-bar (see Fig. 38) 8 feet long and supported at each
I j =^j end, bears a uniform load of 1,200

pounds. The moment of inertia of its
cross -section with respect to the neu-
"<0 tral axis being 2.42 inches*, compute
w the maximum tensile and compressive
-*- unit-stresses in the beam

Evidently the dangerous section

is in the middle, and the value of the maximum bending moment
(see Table B, page 53, Part I) is J WZ, W and I denoting the load
and length respectively. Here

— Wl = -g- X 1,200 X 8 = 1,200 foot-pounds,

or 1,200 X 12 = 14,400 inch-pounds.

The section modulus equals 2.42 -r- 2.28 = 1.06; hence

S = ' „ = 13,585 pounds per square inch.

This is the unit-fibre stress on the lowest fibre at the middle sec-
tion, and hence is tensile. On the highest fibre at the middle
section the unit-stress is compressive, and equals (see page 60):

S' = — S = ^-ott X 13,585 = 4,290 pounds per square inch.

STRENGTH OF MATERIALS 63

EXAMPLES FOR PRACTICE.

1. A beam 12 feet long and 6 X 12 inches in cross-section
rests on end supports, and sustains a load of 3,000 pounds in the
middle. Compute the greatest tensile and compressive unit-
stresses in the beam, neglecting the weight of the beam.

Ans. 750 pounds per square inch.

2. Solve the preceding example taking into account the
weight of the beam, 300 pounds

Ans. 787.5 pounds per square inch.

3. Suppose that a built-in cantilever projects 5 feet from the
wall and sustains an end load of 250 pounds. The cross-section of
the cantilever being represented in Fig. 38, compute the greatest
tensile and compressive unit-stresses, and tell at what places they
occur. (Neglect the weight.)

j Tensile, 4,471 pounds per square inch.

( Compressive, 14,150 " « " "

4. Compute the greatest tensile and compressive unit-stresses
in the beam of Fig. 18, a, due to the loads and the weight of beam
(400 pounds). (A moment diagram is represented in Fig. 18, 5;
for description see example 2, Art. 44, p. 39.) The section of
the beam is a rectangle 8 X 12 inches.

Ans. 580 pounds per square inch.

5. Compute the greatest tensile and compressive unit-stresses
in the cantilever beam of Fig. 19, #, it being a steel I-beam whose
section modulus is 20.4 inches3. (A bending moment diagram for
it is represented in Fig. 19, I; for description, see Ex. 3, Art. 44.)

Ans. 11,470 pounds per square inch.

6. Compute the greatest tensile and compressive unit-stresses
in the beam of Fig. 10, neglecting its weight, the cross-sections
being rectangular 6 X 12 inches. (See example for practice 1,
Art. 43.)

Ans. 600 pounds per square inch.

65. Second Application. The dimensions and the work-
ing strengths of a beam are given, and it is required to determine
its safe load (the manner of application being given).

This problem can be solved by means of equation 6 written
in this form,

M= (6")

64 STRENGTH OF MATERIALS

"We substitute for S the given working strength for the ma-
terial of the beam, and for I and c their values as computed from
the given dimensions of the cross -section; then reduce, thus
obtaining the value of the safe resisting moment of the beam,
which equals the greatest safe bending moment that the beam can
stand. "We next compute the value of the maximum bending
moment in terms of the unknown load; equate this to the value
of the resisting moment previously found; and solve for the
unknown load.

In cast iron, the tensile and compressive strengths are very
different; and the smaller (the tensile) should always be used if
the neutral surface of the beam is midway between the top and
bottom of the beam; but if it is unequally distant from the top
and bottom, proceed as in example 4, following.

Examples. 1. A wooden beam 12 feet long and 6 X 12
inches in cross -section rests on end supports. If its working
strength is 800 pounds per square inch, how large a load uniformly
distributed can it sustain ?

The section modulus is \ba?, b and a denoting the base and
altitude of the section (see Table A, page 52); and here

i la9 =r i x 6 X 122 = 144 inches3.

0 0

Hence S — = 800 X 144 = 115,200 inch-pounds.

For a beam on end supports and sustaining a uniform load, the
maximum bending moment equals JW7 (see Table B, page 55),
"W denoting the sum of the load and weight of beam, and I the
length. If W is expressed in pounds, then

g- Wl = g W X 12 foot-pounds = -^ W X 144 inch-pounds.

Hence, equating tl>Q two values of. maximum bending moment
and the safe resisting moment, we get

lw X 144 — 115,200;

W-.11B 8- 6,400 pounds.

STRENGTH OF MATERIALS 65

The safe load for the beam is 6,400 pounds minus the weight of
the beam.

2. A steel I-beam whose section modulus is 20.4 inches8
rests on end supports 15 feet apart. Neglecting the weight of the
beam, how large a load may be placed upon it 5 feet from one end,
if the working strength is 16,000 pounds per square inch?

The safe resisting moment is

- = 16,000 X 20.4 = 326,400 inch-pounds;
c

hence the bending moment must not exceed that value. The
dangerous section is under the load ; and if P denotes the unknown
value of the load in pounds, the maximum moment (see Table B,
page 53, Part I) equals f P X § foot-pounds, or f P X 60 inch-
pounds. Equating values of bending and resisting moments,
we get

| P X 60 = 326,400;
o

326,400 X 3
or, P = 2'x 6Q = 8,160 pounds.

,3. In the preceding example, it is required to take into
account the weight of the beam. 375 pounds.

5'-

W= 3751105.
Fig. 39.

As we do not know the value of the safe load, we cannot con-
struct the shear diagram and thus determine where the dangerous
section is. But in cases like this, where the distributed load (the
weight) is small compared writh the concentrated load, the dan-
gerous section is practically always where it is under the concen-
trated load alone; in this case, at the load. The reactions due to
the weight equal £ X 375 = 187.5; and the reactions due to the
load equal A P and § P, P denoting the value of the load. The

1. O O " O

larger reaction Rt (Fig. 39) hence equals § P + 187.5. Since

66 STRENGTH OF MATERIALS

the weight of the beam per foot is 375 -j- 15 = 25 pounds, the
maximum bending moment (at the load) equals

( | P + 187.5) 5 - (25 X 5) 2J =

^ P + 937.5 - 312.5 = ^ P + 625.

This is in foot-pounds if P is in pounds.

The safe resisting moment is the same as in the preceding
illustration, 326,400 inch-pounds; hence

(-y P + 625) 12 = 326,400.
Solving for P, we have

, 10 P — 79,725;
or, P = 7,972.5 pounds.

It remains to test our assumption that the dangerous section
is at the load. This can be done by computing Ex (with P =
7,972.5), constructing the shear diagram, and noting where the
shear changes sign. It will be found that the shear changes sign
at the load, thus verifying the assumption.

4. A cast-iron built-in cantilever beam projects 8 feet from
the wall. Its cross-section is represented in Fig. 40, and the

moment of inertia with respect to
the neutral axis is 50 inches4; the
working strengths in tension and
compression are 2,000 and 9,000
pounds per square inch respect-
ively. Compute the safe uniform
load which the beam can sustain,
neglecting the weight of the bearn.

The beam being convex up, the upper fibres are in tension
and the lower in compression. The resisting moment (SI -r- c),
as determined by the compressive strength, is

STRENGTH OF MATERIALS 67

= 100,000 inch-pounds;

and the resisting moment, as determined by the tensile strength, is

2,000 X 50

- ~-~ - = 40.000 inch-pounds.

6.U

Hence the safe resisting moment is the lesser of these two, or
40,000 inch-pounds. The dangerous section is at the wall (see
Table B, page 53), and the value of the maximum bending
moment is \ WZ, W denoting the load and I the length. If W is
in pounds, then

M = i W X 8 foot-pounds = £ W X 96 inch-pounds.
Equating bending and resisting moments, we have

-i-W X 96 = 40,000;

w= 40,000 X 2 ^833poundSt

EXAMPLES FOR PRACTICE.

1. An 8 X 8 -inch timber projects 8 feet from a wall. If its
working strength is 1,000 pounds per square inch, how large an
end load can it safely sustain ?

Ans. 890 pounds.

2. A beam 12 feet long and 8 X 16 inches in cross -section,
on end supports, sustains two loads P, each 3 feet from its ends
respectively. The working strength being 1,000 pounds per square
inch, compute P (see Table B, page 53).

Ans. 9,480 pounds.

3. An I-beam weighing 25 pounds per foot rests on end
supports 20 feet apart. Its section modulus is 20.4 inches3, and
its working strength 16,000 pounds per square inch. Compute
the safe uniform load which it can sustain.

Ans. 10,880 pounds-

66. Third Application. The loads, manner of support,
and working strength of beam are given, and it is required to de-
termine the size of cross-section necessary to sustain the load
safely, that is, to "design the beam."

68 STRENGTH OF MATERIALS

To solve this problem, we use the first beam formula (equation
6), written in this form,

JL *L. (6'")

o ' S

We first determine the maximum bending moment, and then sub-
stitute its value for M, and the working strength for S. Then we
have the value of the section modulus (I -r- <?„) of the required
beam. Many cross-sections can be designed, all having a given
section modulus. Which one is to be selected as most suitable will
depend on the circumstances attending the use of the beam and
on considerations of economy.

.Examples. 1. A timber beam is to be used for sustaining
a uniform load of 1,500 pounds, the distance between the supports
being 20 feet. If the working strength of the timber is 1,000 pounds
per square inch, what is the necessary size of cross-section ?

The dangerous section is at the middle of the beam; and the
maximum bending moment (see Table B, page 53) is

-i-WJ = ~ x 1,500 X 20 = 3,750 foot-pounds,

or 3,750 X 12 = 45,000 inch-pounds.

I 45,000

Hence • — = ., AAA = 45 inches3.

c 1,UUU

Now the section modulus of a rectangle is \bo? (see Table A,
page 54, Part I); therefore, \la? = 45, or be? = 270.

Any wooden beam (safe strength 1,000 pounds per square
inch) whose breadth times its depth square equals or exceeds 270,
is strong enough to sustain the load specified, 1,500 pounds.

To determine a size, we may choose any value for 5 or a, and
solve the last equation for the unknown dimension. It is best,
however, to select a value of the breadth, as 1, 2, 3, or 4 inches,
and solve for a. Thus, if we try b = 1 inch, we have

a2 = 270, or a = 16.43 inches.

This would mean a board 1 X 18 inches, which, if used, would
have to be supported sidewise so as to prevent it from tipping or
" buckling." Ordinarily, this would not be a good size.

Next try J = 2 inches; we have

2 x a2 = 270; or a — 1/270 -f- 2 = 11.62 inches.
This would require a plank 2 X 12, a better proportion than the
first. Trying 5 = £ inches, we have

STRENGTH OF MATERIALS 69

3 x a? = 270; or a = 1/270 -*- 3 = 9.49 inches.
This would require a plank 3 X 10 inches; and a choice between
a 2 X 12 and a 3 X 10 plank would be governed by circumstances
in the case of an actual construction.

It will be noticed that we have neglected the weight of the
beam. Since the dimensions of wooden beams are not fractional,
and we have to select a commercial size next larger than the one
computed (12 inches instead of 11.62 inches, for example), the
additional depth is usually sufficient to provide strength for the
weight of the beam. If there is any doubt in the matter, we can
settle it by computing the maximum bending moment including
the weight of the beam, and then computing the greatest uni*-fibre
stress due to load and weight. If this is less than the safe strength,
the section is large enough; if greater, the section is too small.

Thus, let us determine whether the 2 X 12-inch plank is
strong enough to sustain the load and its own weight. Tb« plank
will weigh about 120 pounds, making a total load of

1,500 + 120 = 1,620 pounds.
Hence the maximum bending moment is

JlWZ == 4-1,620 X 20 X 12 = 48,600 inch-pounds.

o o

Since -i = i l>a2 = -^-X 2 X 122 = 48, and S = j^L,

S = - ' = 1,013 pounds per square inch.

Strictly, therefore, the 2 X 12-inch plank is not large enough; but
as the greatest unit-stress in it would be only 13 pounds per square
inch too large, its use would be permissible.

2. What size of steel I-beam is needed to sustain safely the
loading of Fig. 9 if the safe strength of the steel is 16,000 pounds
per square inch ?

The maximum bending moment due to the loads was found
in example 1, Art. 43, to be 8,800 foot-pounds, or 8,800 X 12 =
105,600 inch-pounds.

I 105,600
— T =

That is, an I-beam is needed whose section modulus is a little
larger than 6.6, to provide strength for its own weight.

STRENGTH OF MATERIALS

To select a size, we need a descriptive table of I-beams, such
as is published in handbooks on structural steel.

Below is an abridged copy of such a table. (The last two columns con-
tain information for use later.) The figure illustrates a cross-section of an
I-beam, and shows the axes referred to in the table.

It will be noticed that two sizes are given for each depth;
these are the lightest and heaviest of each size that are made, but
intermediate sizes can be secured. In column 5 we find 7. 3 as the
next larger section modulus than the one required (6,6); and this
corresponds to a 12^-pound 6-inch I-beam, which is probably the
proper size. To ascertain whether the excess (7.3-6.6 = 0.70)
in the section modulus is sufficient to provide for the weight of the
beam, we might proceed as in example 1. In this case, however,
the excess is quite large, and the beam selected is doubtless safe.

TABLE C.

Properties ot Standard I-Beams

Section of beam, showing axes 1-1 and 2-2.

Depth of
Beam,
in inches.

Weight
per foot,
in pounds.

Area of cross-
section, in
square inches.

Moment of
inertia,
axis 1—1.

Section
modulus,
axis 1—1.

Moment of
inertia,
axis 2—2.

5.50

1.63

2.5

1.7

0.46

7.50

2.21

2.9

1.9

.60

7.50

2.21

6.0

3.0

.77

10.50

3.09

7.1

3.6

1.01

9.75

2.87

12.1

4.8

1.23

14.75

4.34

15.1

6.1

1.70

12.25

3.61

21.8

7.3

1.85

17.25

£.07

26.2

8.7

2.36

15.00

4.42

36.2

10.4

2.67

20.00

5.88

42.2

12.1

3.24

18.00

5.33

56.9

14.2

3.78

25.25

7.43

68.0

17.0

4.71

21.00

6.31

84.9

18.9

5.16

35.00

10.29

111.8

24.8

7.31

25.00

7.37

122.1

24.4

6.89

40.00

11.76

158.7

31.7

9.50

31.50

9.26

215.8

36.0

9.50

40.00

11.76

245.9

41.0

10.95

42.00

12.48

441.8

58.9

14.62

60.00

17.65

538.6

71.8

18.17

55.00

15.93

795.6

88.4

21.19

70.00

20.59

921.2

102.4

24.62

65.00

19.08

1,169.5

117.0

27.86

75.00

22.06

1,268.8

126.9

30.25

80.00

23.32

2,087.2

173.9

42.86

100.00

29.41

2,379.6

198.3

48.55

STRENGTH OF MATERIALS 71

EXAMPLES FOR PRACTICE.

1. Determine the size of a wooden beam which can safely
sustain a middle load of 2,000 pounds, if the beam rests on end
supports 16 feet apart, and its working strength is 1,000 pounds
per square inch. Assume width 6 inches.

Ans. 6 X 10 inches.

2. What sized steel I-beam is needed to sustain safely a
uniform load of 200,000 pounds, if it rests on end supports 10
feet apart, and its working strength is 16,000 pounds per square
inch?

Ans. 95-pound 24-incli.

3. What sized steel I-beam is needed to sustain safely the
loading of Fig. 10, if its working strength is 16,000 pounds per
square inch ?

Ans. 14.75-pound 5 -inch.

67. Laws of Strength of Beams. The strength of a beam is
measured by the bending moment that it can safely withstand ; or,
since bending and resisting moments are equal, by its safe resist,
ing moment (SI -f- c). Hence the safe strength of a beam varies
(1) directly as the working fibre strength of its material, azid (2)
directly as the section modulus of its cross-section. For beams
rectangular in cross-section (as wooden beams), the section modu-
lus is \bc?, b and a denoting the breadth and altitude of the
rectangle. Hence the strength of such beams varies also directly
as the breadth, and as the square of the depth. Thus, doubling
the breadth of the section for a rectangular beam doubles the
strength, but doubling the depth quadruples the strength.

The safe load that a beam can sustain varies directly as its
resisting moment, and depends on the way in which the load is
distributed and how the beam is supported. Thus, in the first
four and last two cases of the table on page 55,

M = PZ, hence P =-- SI -f- Ic,

M = £ Wl, " W = 2SI -r- Ic,

M = | P/, " P = 4SI -*• Ic,

M = £ WZ, « W = SSI -4- Ic,

M = I PI, « P = 881 -f- Ic,

M = ^ WZ, « W = 12SI -*- fo.

STRENGTH OF MATERIALS

Therefore the safe load in all cases varies inversely with the
length; and for the different cases the safe loads are as 1, 2, 4, 8,
8, and 12 respectively.

Example. What is the ratio of the strengths of a plank 2 X
10 inches when placed edgewise and when placed flatwise on its
supports ?

When placed edgewise, the section modulus of the plank is
\ X 2 X 102=-- 33 J, and when placed flatwise it is \ X 10 X 22 =
6J-; hence its strengths in the two positions are as 33J to 6-|
respectively, or as 5 to 1.

EXAMPLE FOR PRACTICE.

What is the ratio of the safe loads for two beams of wood,
one being 10 feet long, 3x12 inches in section, and having its load
in the middle; and the other 8 feet long and 2x8 inches in section,
with its load uniformly distributed.

Ans. As 28.8 to 21. 3

68. Modulus of Rupture. If a beam is loaded to destruction,
and the value of the bending moment for the rupture stage is
computed and substituted for M in the formula SI -r- c = M, then
the value of S computed from the equation is the modulus of
rupture for the material of the beam. Many experiments have
been performed to ascertain the moduli of rupture for different
materials and for different grades of the same material. The fol-
owing are fair values, all in pounds per square inch :

TABLE D.

Moduli of Rupture.

Timber:
Spruce

4,000— 7,000, aver
3,500 7,000,
5,500 10,500,
10,000 16,000,
8,000 14,000,
4,000 12,000,
7,500 18,500,
9,000 15,000,

age 5,000
4,500
8,000
12,500
10,000
8,000
13,000
11,500

Hemlock .

White pine.

Long-leaf pine. . . .
Short-leaf pine. . .
Douglas spruce. . .
White oak .

Red oak

Stone :
Sandstone.

400— 1,200,
400 1,000.
800 1,400.

Limestone.

Granite

Cast iron:

One and one-half to two and
one-quarter times its ulti-
mate tensile strength.

Hard steel:

Varies from 100,000 to 150,000

STRENGTH OF MATERIALS 73

Wrought iron and structural steels have no modulus of rup-
ture, as specimens of those materials will " bend double," but not
break. The modulus of rupture of a material is used principally
as a basis for determining its working strength. The factor of
safety of a loaded beam is computed by dividing the 'modulus
of rupture of its material by the greatest unit --fibre stress in
the beam.

69. The Resisting 5hear. The shearing stress on a cross-
section of a loaded beam is not a uniform stress; that is, it is not
uniformly distributed over the section. In fact the intensity or
unit-stress is actually zero on the highest and lowest fibres of a
cross-section, and is greatest, in such beams as are used in prac-
tice, on fibres at the neutral axis. In the following article we
explain how to find the maximum value in two cases — cases which
are practically important.

70. Second Beam Formula. Let Ss denote the average
value of the unit-shearing stress on a cross-section of a loaded
beam, and A the area of the cross-section. Then the value of the
whole shearing stress on the section is :

Resisting shear = Ss A.

Since the resisting shear and the external shear at any section of a
beam are equal (see Art. 59),

SSA = V. (7)

This is called the " second beam formula " It is used to investi-
gate and to design for shear in beams.

In beams uniform in cross-section, A is constant, and Ss is
greatest in the section for which Y is greatest. Hence the great-
est unit-shearing stress in a loaded beam is at the neutral axis of
the section at which the external shear is a maximum. There is
a formula for computing this maximum value in any case, but it
is not simple, and we give a simpler method for computing the
value in the two practically important cases:

1. In wooden beams (rectangular or square in cross-section), the
greatest unit-shearing stress in a section is 50 per cent larger than the average
value S§.

2. In I-beams, and in others with a thin vertical web, the greatest
unit-shearing stress in a section practically equals S8, as given by equation 7,
if the area of the web is substituted for A.

74 STRENGTH OF MATERIALS

Examples. 1. What is the greatest value of the unit-
shearing stress in a wooden beam 12 feet long and 6x12 inches in
cross-section when resting on end supports and sustaining a uni-
form load of 6,400 pounds ? (This is the safe load as determined
by working fibre stress; see example 1, Art. 65.)

The maximum external shear equals one-half the load (see
Table B, page 53), and comes on the sections near the supports.

Since A = 6 X 12 = 72 square inches;

3,200
s ~~ 72 = pounds per square inch,

and the greatest unit- shearing stress equals

3 3

~2~ ^s = ~2~ ^ = ^ pounds per square inch.

Apparently this is very insignificant; but it is not negligible, as
is explained in the next article.

2. A steel I-beam resting on end supports 15 feet apart
sustains a load of 8,000 pounds 5 feet from one end. The weight
of the beam is 375 pounds, and the area of its web section is 3.2
square inches. (This is the beam and load described in examples
2 and 3, Art. 65.) What is the greatest unit-shearing stress ?

The maximum external shear occurs near the support where
the reaction is the greater, and its value equals that reaction.
Calling that reaction R, and taking moments about the other end
of the beam, we have

R x 15 - 375 x 7-7T - 8,000 x 10 = 0;

therefore 15 R = 80,000 + 2,812.5 === 82,812.5;
or, R = 5,520.8 pounds.

Hence Sg = ' — = 1,725 pounds per square inch.

EXAMPLES FOR PRACTICE.

1. A wooden beam 10 feet long and 2 X 10 inches in cross-
section sustains a middle load of 1,000 pounds. Neglecting the
weight of the beam, compute the value of the greatest unit-shearing

stress.

Ans. 37.5 pounds per square inch.

STRENGTH OF MATERIALS 75

2. Solve the preceding example taking into account the
weight of the beam, 60 pounds.

Ans. 40 pounds per square inch. v

3. A wooden beam 12 feet long and 4 X 12 inches in cross-
section sustains a load of 3,000 pounds 4 feet from one end.
Neglecting the weight of the beam, compute the value of the
greatest shearing unit-stress.

Ans. G2.5 pounds per square inch.

71. Horizontal Shear. It can be proved that there is a
shearing stress on every horizontal section of a loaded beam. An
experimental explanation will have to suffice here. Imagine a
pile of six boards of equal length supported so that they do not
bend. If the intermediate supports are removed, they will bend
and their ends will not be flush but somewhat as represented in
Fig. 41. This indicates that the boards slid over each other during
the bending, and hence there was a rubbing and a frictional re-
sistance exerted by the boards upon each other. Now, when a
solid beam is being bent, there is an exactly similar tendency for
the horizontal layers to slide over each other; and, instead of a
frictional resistance, there exists shearing stress on all horizontal
sections of the beam.

In the pile of boards the amount of slipping is different at
different places between any two boards, being greatest near the
,supports and zero midway between them. Also, in any cross-
section the slippage is least between the upper two and lower two
boards, and is greatest between the middle two. These facts indi-
cate that the shearing unit- stress on horizontal sections of a solid
beam is greatest in the neutral surface at the supports.

It can be proved that at any place in a beam the shearing
unit-stresses on a horizontal and on a vertical section are equal.

Fig. 41. Fig. 42.

It follows that the horizontal shearing unit- stress is greatest at the
neutral axis of the section for which the external shear (V) is a
maximum. Wood being very weak in shear along the grain,
timber beams sometimes fail under shear, the "rupture" being

76 STRENGTH OF MATERIALS

two horizontal cracks along the neutral surface somewhat as rep-
resented in Fig. 42. It is therefore necessary, when dealing with
timber beams, to give due attention to their strength as determined
by the working strength of the material in shear along the grain.

Example. A wooden beam 3 X 10 inches in cross-section
rests on end supports and sustains a uniform load of 4,000 pounds
Compute the greatest horizontal unit-stress in the beam.

The maximum shear equals one-half the load (see Table B,
page 55), or 2,000 pounds. Hence, by equation 7, since A =
3 X 10 = 30 square inches,

2,000 2

s = 30 =6677- pounds per square inch.

This is the average shearing unit-stress on the cross-sections near
the supports; and the greatest value equals

3 2

- X 66-7- = 100 pounds per square inch.

According to the foregoing, this is also 1?he value of the
greatest horizontal shearing unit-stress. (If of white pine, for
example, the beam would not be regarded as safe, since the ulti-
mate shearing strength along the grain of selected pine is only
about 400 pounds per square inch.)

72. Design of Timber Beams. In any case we may pro-
ceed as follows: — (1) Determine the dimensions of the cross-
section of the beam from a consideration of the fibre stresses as'
explained in Art. 66. (2) With dimensions thus determined, com-
pute the value of the greatest shearing unit-stress from the formula,

Greatest shearing unit-stress — -^- Y -=- ab,

where Y denotes the maximum external shear in the beam, and
b and a the breadth and depth of the cross-section.

If the value of the greatest shearing unit-stress so computed
does not exceed the working strength in shear along the grain,
then the dimensions are large enough; but if it exceeds that value,
then a or &, or both, should be increased until J. V -s- ab is less
than the working strength. Because timber beams are very often
"season checked" (cracked) along the neutral surface, it is ad vis-

STRENGTH OF MATERIALS 77

able to take the working strength of wooden beams, in shear along
the grain, quite low. One-twentieth of the working fibre strength
has been recommended* for all pine beams.

If the working strength in shear is taken equal to one-
twentieth the working fibre strength, then it can be shown that,

1. For a beam on end supports loaded in the middle, the safe load de-
pends on the shearing or fibre strength according as the ratio of length to
depth (I -s- a) is less or greater than 10.

2. For a beam on end supports uniformly loaded, the safe load depends
on the shearing or fibre strength according as I + a is less or greater than 20.

Examples. 1. It is required to design a timber beam to sus-
tain loads as represented in Fig. 11, the working fibre strength
being 550 pounds and the working shearing strength 50 pounds
per square inch.

The maximum bending moment (see example for practice 3,
Art. 43; and example for practice 2, Art. 44) equals practically
7,000 foot-pounds or, 7,000 X 12 = 84,000 inch-pounds.
Hence, according to equation 6'",

I 84,000

= -sra- = 152.7 inches8.
c 550

Since for a rectangle

- ba* = 152.7, or ba* = 916.2.

Now, if we let 5 = 4, then a2 = 229;

or, a = 15.1 (practically 16) inches.

If, again, we let b — 6, then a2 = 152.7;

or a = 12.4 (practically 14) inches.

Either of these sizes wrill answer so far as fibre stress is concerned,
but there is more " timber " in the second.

The maximum external shear in the beam equals 1,556
pounds, neglecting the weight of the beam (see example 3, Art.
87; and example 2, Art. 38). Therefore, for a 4 X 16-inch beam,

* See "Materials of Construction." — JOHNSON. Page 55.

78 STRENGTH OF MATERIALS

3 ; . 1,556
Greatest shearing unit- stress =-gr x 4 v ifi

= 36.5 pounds per square inch;
and for a 6 X 14-inch beam, it equals

3 1,556
~9~ X ^ - TJ = 27.7 pounds per square inch.

Since these values are less than the working strength in shear,
either size of beam is safe as regards shear.

If it is desired to allow for weight of beam, one of the sizes
should be selected. First, its weight should be computed, then
the new reactions, and then the unit-fibre stress may be com-
puted as in Art. 64, and the greatest shearing unit-stress as in the
foregoing. If these values are within the working values, then
the size is large enough to sustain safely the load and the weight
of the beam.

2. What is the safe load for a white pine beam 9 feet long
and 2x12 inches in cross-section, if the beam rests on end supports
and the load is at the middle of the beam, the working fibre
strength being 1,000 pounds and the shearing strength 50 pounds
per square inch.

The ratio of the length to the depth is less than 10; hence
the safe load depends on the shearing strength of the material
Calling the load P, the maximum external shear (see Table B,
page 53) equals -J- P, and the formula for greatest shearing unit
stress becomes

3 -i- P

50 = --x 2; or P = lj600 Pounds-

EXAMPLES FOR PRACTICE.

1. "What size of wooden beam can safely sustain loads as in
Fig. 12T with shearing and fibre working strength equal to 50 and
1,000 pounds per square inch respectively ?

f:.. Ans. 6 X 12 inches

2. What is the safe load for a wooden beam 4 X 14 inches,
and 18 feet long, if the beam rests on end supports and the load
is uniformly distributed, with working strengths as in example 1 ?

Ans. 3,730 pounds

STRENGTH OF MATERIALS 79

73. Kinds of Loads and Beams. We shall now discuss the
strength of beams under longitudinal forces (acting parallel to
the beam) and transverse loads. The longitudinal forces are
supposed to be applied at the ends of the beams and along the axis*
of the beam in each case. We consider only beams resting on
end supports.

The transverse forces produce bending or flexure, and the
longitudinal or end forces, if pulls, produce tension in the beam;
if pushes, they produce compression. Hence the cases to be con-
sidered may be called " Combined Flexure and Tension " and
" Combined Flexure and Compression."

74. Flexure and Tension. Let Fig. 43, <z, represent a beam
subjected to the transverse loads L15 L2 and L3, and to two equal
end pulls P and P. The reactions Rx and R2 are due to the trans-
verse loads and can be computed by the methods of moments just
as though there were no end pulls. To find the stresses at any
cross -section, we determine those due to the transverse forces
(Lj, L2, L3, R1 and R2) and those due to the longitudinal; then
combine these stresses to get the total effect of all the applied
forces.

The stress due to the transverse forces consists of a shearing
stress and a fibre stress; it will be called the flexural stress. The
fibre stress is compressive above and tensile below. Let M denote
the value of the bending moment at the section considered; cl and
c2 the distances from the neutral axis to the highest and the low-
est fibre in the section ; and Sj and S2 the corresponding unit-fibre
stresses due to the transverse loads. Then

, 2

bx == — y— ; and !32 = -y- .

The stress due to the end pulls is a simple tension, and it equals
P; this is sometimes called the direct stress. Let S0 denote the
unit-tension due to P, and A the area of the cross-section; then

a P
b° " A-

Both systems of loads to the left of a section between Lj and

* NOTE. By " axis of a beam " is meant the line through the centers of
gravity of all the cross-sections.

80 STRENGTH OF MATERIALS

L2 are represented in Fig. 43, &; also the stresses caused by them
at that section. Clearly the effect of the end pulls is to increase the
tensile stress (on the lower
fibres) and to decrease the
compressive stress (on the
upper fibres) due to the flex-
u^e. Let Sc denote the total
(resultant) unit- stress on the
>per fibre, and St that on
e lower fibre, due to all
'\ the forces acting on the beam.
In combining the stresses
there are two cases to con-
sider:

(1) The flexural compressive unit- stress on the upper fibre is
greater than the direct unit-stress; that is, Sj is greater than S0.
The resultant stress on the upper fibre is

Sc = Sj - S0 (compressive) ;
and that on the lower fibre is

St = S2 + S0 (tensile).

The combined stress is as represented in Fig. 43, c, part tensile
and part compressive.

(2) The flexural compressive unit-stress is less than the
direct unit-stress; that is, Sj is less than S0. Then the combined
unit-stress on the upper fibre is

Sc = 80-8! (tensile);
and that on the lower fibre is

St = S2 + S0 (tensile).

The combined stress is represented by Fig. 43, d, and is all
tensile.

Example. A steel bar 2x6 inches, and 12 feet long, is sub-
jected to end pulls of 45,000 pounds. It is supported at each
end, and sustains, as a beam, a uniform load of 6,000 pounds.
It is required to compute the combined unit-fibre stresses.

Evidently the dangerous section is at the middle, and M =

• ; that is,

STRENGTH OF MATERIALS 81

M = — X 6,000 X 12 = 9,000 foot-pounds,

or 9,000 X 12 = 108,000 inch-pounds.

The bar being placed with the six-inch side vertical,
c1 = c2 = 3 inches, and
I=JLx2x63 = 36 inches4. (See Art. 52.)

I/O

108,000 X 3

Hence S, = S2 = —^ — - = 9,000 pounds per square inch.

Since A = 2 X 6 = 12 square inches,

45,000

fe0 = — ^— — = o,7oU pounds per square men.
l<o

The greatest value of the combined compressive stress is

S, - S0 = 9,000 - 3,750 = 5,250 pounds per square inch,
and it occurs on the upper fibres of the middle section. The great-
est value of the combined tensile stress is

S2 + S0 = 9,000 + 3,750 = 12,750 pounds per square inch,
and it occurs on the lowest fibres of the middle section.

EXAHPLE FOR PRACTICE.

Change the load in the preceding illustration to one of 6,000
pounds placed in the middle, and then solve.

A ( Sc = 14,250 pounds per square inch.
QS' j St = 21,750 « « « «

75. Flexure and Compression. Imagine the arrowheads on
P reversed; then Fig. 43, «, will represent a beam under com-
bined flexural and compressive stresses. The flexural unit- stresses
are computed as in the preceding article. The direct stress is a
compression equal to P, and the unit-stress due to P is computed
as in the preceding article. Evidently the effect of P is to increase
the compressive stress and decrease the tensile stress due to the
flexure. In combining, we have two cases as before:

(1) The flexural tensile unit-stress is greater than the
direct unit-stress; that is, S2 is greater than S0. Then the com-
bined unit-stress on the lower fibre is

82 STRENGTH OF MATERIALS

St = S2 - S0 (tensile) ;
and that on the upper fibre is

Sc = S, -f S0 (compressive).

The combined fibre stress is represented by Fig. 44, a, and is part
tensile and part compressive.

(2) The flexural unit-stress on the lower fibre is less than
the direct unit-stress; that is, S2 is less than S0. Then the com-
bined unit-stress on the lower fibre is

St = S0 - S2 (compressive);
and that on the upper fibre is

Sc = S0 + Sj (compressive).
The combined fibre stress is represented by
Fig. 44, £, and is all compressive.

Example. A piece of timber 6x6
inches, and 10 feet long, is subjected to end
pushes of 9,000 pounds. It is supported in
a horizontal position at its ends, and sustains
a middle load of 400 pounds. Compute the
combined fibre stresses.

Evidently the dangerous section is at the
middle, and M = l P/; that is, Fig. 44.

M = j X 400 X 10 = 1,000 foot-pounds,

or 1,000 X 12 — 12,000 inch-pounds.

Since cl = c2 = 3 inches, and

1 3 1

1 r= "To k># == ~TT\ X o X 6 = 108 inches4,
l/o ±/4

12,000 X 3 1

bx — fe2 _ ^Qg = ~3~ pounds per square inch,

Since A = 6 X 6 = 36 square inches,

9 000
S0 = ' = 250 pounds per square inch.

Hence the greatest value of the combined compressive stress is
So + Sa = 333 -g- -f 250 = 583-q- pounds per square inch.

STRENGTH OF MATERIALS

Ans.

It occurs on the upper fibres of the middle section. The greatest
value of the combined tensile stress is

S2 - S0 = 333-g- - 250 = 83-^- pounds per square inch.
It occurs on the lowest fibres of the middle section.

EXAMPLE FOR PRACTICE.

Change the load of the preceding illustration to a uniform
load and solve.

Sc = 417 pounds per square inch.

St == 83 « « « " (compression).

76. Combined Flexural and Direct Stress by flore Exact
Formulas. The results in the preceding articles are only approxi-
mately correct. Imagine the
beam represented in Fig. 45, a,
to be first loaded with the trans-
verse loads alone. They cause
the beam to bend more or less,
and produce certain flexural
stresses at each section of the
beam. Now, if end pulls are
applied they tend to straighten
the beam and hence diminish the flexural stresses. This effect
of the end pulls wTas omitted in the discussion of Art. 74, and
the results there given are therefore only approximate, the
value of the greatest combined fibre unit-stress (St) being too
large. On the other hand, if the end forces are pushes, they in-
crease the bending, and therefore increase the flexural fibre stresses
already caused by the transverse forces (see Fig. 45, b). The
results indicated in Art. 75 must therefore in this case also be
regarded as only approximate, the value of the greatest unit-
fibre stress (Sc) being too small.

For beams loaded in the middle or with a uniform load, the
following formulas, which take into account the flexural effect of
the end forces, may be used :

M denotes bending moment at the middle section of the beam'
I denotes the moment of inertia of the middle section with
respect to the neutral axis;

84 STRENGTH OF MATERIALS

S1? S2, c1 and c2 have the same meanings as in Arts. 74 and
75, but refer always to the middle section ;

I denotes length of the beam ;

E is a number depending on the stiffness of the material, the
average values of which are, for timber, 1,500,000; and for struc-
tural steel 30,000,000.*

~ 10E

The plus sign is to be used when the end forces P are pulls, and
the minus sign when they are pushes.

It must be remembered that St and S2 are flexural unit-
stresses. The combination of these and the direct unit-stress is
made exactly as in articles 74 and 75.

Examples. 1. It is required to apply the formulas of this
article to the example of article 74.

As explained in the example referred to, M = 108,000 inch-
pounds; Cj= c2= 3 inches; and I = 36 inches4.
Now, since I = 12 feet = 144 inches,

108,000 X 3 324,000

8' = S* = 45,000 X 144* = 36 + Ml == 8'284 P°unda

^ 10 X 30,000,000

per square inch, as compared with 9,000 pounds per square inch,
the result reached by the use of the approximate formula.
As before, S0 — 3,750 pounds per square inch; hence

Sc = 8,284- 3,750 = 4,534 pounds per square inch;
and St = 8,284 + 3,750 = 12,034 " " " «

2. It is required to apply the formulas of this article to the
example of article 75.

As explained in that example,

M = 12,000 inch-pounds;
c1 = <?2 =. 3 inches, and I = 108 inches4.
Now, since I = 120 inches,

12,000 X 3 36,000

• 362

10 X 1,500,000
* NOTE. This quantity " E " is more fully explained in Article 95.

STRENGTH OF MATERIALS 85

per square inch, as compared with 333J pounds per square inch,
the result reached by use of the approximate method.
As before, S0 = 250 pounds per square inch; hence

Sc = 362 -f- 250 = 612 pounds per square inch; and
St == 362 - 250 =±= 112 « « « « .

EXAMPLES FOR PRACTICE.

1. Solve the example for practice of Art. 74 by the formulas
of this article.

A n * $ Sc = 12,820 pounds per square inch.
1S' I St = 20,320 « « «

2. Solve the example for practice of Art. 75 by the formulas
of this article.

Ans i ^c ~ ^^ pounds per square inch.

| St = 70 " « « « (compression).

STRENGTH OF COLUHNS.

A stick of timber, a bar of iron, etc., when used to sustain
end loads which act lengthwise of the pieces, are called columns,
posts, or struts if they are so long that they would bend before
breaking. When they are so short that they would not bend
before breaking, they are called short blocks, and their compres-
sive strengths are computed by means of equation 1. The strengths
of columns cannot, however, be so simply determined, and we now
proceed to explain the method of computing them.

77. End Conditions. The strength of a column depends in
part on the way in which its ends bear, or are joined to other
parts of a structure, that is, on its " end conditions." There are
practically but three kinds of end conditions, namely:

1. " Hinge " or " pin " ends,

2. "Flat" or "square" ends, and

3. "Fixed" ends.

(1) When a column is fastened to its support at one end by
means of a pin about which the column could rotate if the other
end were free, it is said to be " hinged " or " pinned " at the
former end. Bridge posts or columns are often hinged at the ends.

(2) A column either end of which is flat and perpendicular
to its axis and bears on other parts of the structure at that surface,

square" at that end.

STRENGTH OF MATERIALS

(3) Columns are sometimes riveted near their ends directly
to other parts of the structure and do not bear directly on their
ends; such are called " fixed ended." A column which bears on its
flat ends is often fastened near the ends to other parts of the struc-
ture, and such an end is also said to be " fixed." The fixing of an
end of a column stiffens and therefore strengthens it more or less,
but the strength of a column with fixed ends is computed as
though its ends were flat. Accordingly we have, so far as strength
is concerned, the following classes of columns :

78. Classes of Columns. (1) Both ends hinged or pinned;
(2) one end hinged and one flat; (3) both ends flat.

Other things being the same, columns of these three classes
are unequal in strength. Columns of the first class are the
weakest, and those of the third class are the strongest.

_A_

=•

_A_

.A.

Fig. 46.

|B
b

79. Cross=sections of Columns. Wooden columns are usu-
ally solid, square, rectangular, or round in section ; but sometimes
they are " built up " hollow. Cast-iron columns are practically
always made hollow, and rectangular or round in section. Steel
columns are made of single rolled shapes — angles, zees, channels,
etc.; but the larger ones are usually " built up" of several shapes.
Fig. 46, &, for example, represents a cross -section of a " Z-bar"
column; and Fig. 46, &, that of a " channel " column.

80. Radius of Gyration. There is a quantity appearing in
almost all formulas for the strength of columns, which is called
u radius of gyration." It depends on the form and extent of the
cross-section of the column, and may be defined as follows:

STRENGTH OF MATERIALS 87

The radius of gyration of any plane figure (as the section of a column)
with respect to any line, is such a length that the square of this length mul-
tiplied by the area of the figure equals the moment of inertia of the figure
with respect to the given line.

Thus, if A denotes the area of a figure; I, its moment of in-
ertia with respect to some line; and r, the radius of gyration
with respect to that line; then

r*A. = I',orr = I/I H- A. (9)

In the column formulas, the radius of gyration always refers to an
axis through the center of gravity of the cross-section, and usually
to that axis with respect to which the radius of gyration (and mo.
ment of inertia) is least. (For an exception, see example 3,
Art. 83.) Hence the radius of gyration in this connection is often
called for brevity the " least radius of gyration," or simply the
" least radius."

Examples. 1. Show that the value of the radius of gyration
given for the square in Table A, page 52, is correct.

The moment of inertia of the square with respect to the axis
is TV^*» Since A = #2, then, by formula 9 above,

r - J17I7 = JITa, = a J J.

2. Prove that the value of the radius of gyration given for
the hollow square in Table A, page 54, is correct.

The value of the moment of inertia of the square with respect
to the axis is T^ (a* - a'). Since A = a2 - a*,

EXAHPLE FOR ^PRACTICE.

Prove that the values of the radii of gyration of the other fig-
ures given in Table A, page 52, are correct. The axis in each
case is indicated by the line through the center of gravity.

81. Radius of Gyration of Built=up Sections. The radius of
gyration of a built-up section is computed similarly to that of any
other figure. First, we have to compute the moment of inertia of

88 STRENGTH OF MATERIALS

the section, as explained in Art. 54; and then we use formula 9, as
in the examples of the preceding article.

Example. It is required to compute . the radius of gyration
of the section represented in Fig. 30 (page 52) with respect to the
axis AA.

In example 1, Art. 54, it is shown that the moment of inertia
of the section with respect to the axis AA is 429 inches4. The
area of the whole section is

2 X 6.03 + 7 = 19.06;
hence the radius of gyration r is

f429~
\ 153)6 =

4.74 inches.

EXAMPLE FOR PRACTICE.

Compute the radii of gyration of the section represented in
Fig. 31, #, with respect to the axes AA and BB. (See examples
for practice 1 and 2, Art. 54.)

A ( 2.87 inches.

Ans. <

\ 2.09 «

82. Kinds of Column Loads. When the loads applied to a
column are such that their resultant acts through the center of
gravity of the top section and along the axis of the column, the
column is said to be centrally loaded. When the resultant of the
loads does not act through the center of gravity of the top
section, the column is said to be eccentrically loaded. All the
following formulas refer to columns centrally loaded.

83. Rankine's Column Formula. When a perfectly straight
column is centrally loaded, then, if the column does not bend and
if it is homogeneous, the stress on every cross-section is a uniform
compression. If P denotes the load and A the area of the cross-
section, the value of the unit-compression is P -r- A.

On account of lack of straightness or lack of uniformity in
material, or failure to secure exact central application of the load,
the load P has what is known as an " arm " or " leverage " and
bends the column more or less. There is therefore in such a
column a bending or flexural stress in addition to the direct com-
pressive stress above mentioned ; this bending stress is compressive

STRENGTH OF MATERIALS 89

on the concave side and tensile on the convex. The value of the
stress per unit-area (unit-stress) on the fibre at the concave side,
according to equation 6, is Me -j- I, where M denotes the bending
moment at the section (due to the load on the column), c the
distance from the neutral axis to the concave side, and I the
moment of inertia of the cross-section with respect to the neutral
axis. (Notice that this axis is perpendicular to the plane in
which the column bends.)

The upper set of arrows (Fig. 47) represents the direct com-
pressive stress; and the second set the bending stress if the load
is not excessive, so that the stresses are within the elastic limit of
the material. The third set represents the combined stress that
actually exists on the cross-section. The greatest combined unit-
stress evidently occurs on the fibre at the concave side and where
the deflection of the column is greatest. The
stress is compressive, and its value S per unit-
area is given by the formula,

P Me

= TT T-

Now, the bending moment at the place of
greatest deflection equals the product of the
load P and its arm (that is, the deflection).
^* Calling the deflection d, we have M = P</; and
this value of M, substituted in the last equa
f^r^^^^^^—- tion, gives

Fig. 47.

Let r denote the radius of gyration of the cross-section with respec
to the neutral axis. Then I = Ar2 (see equation 9); and this
value, substituted in the last equation, gives

P Rfo P do

= X + A? - ~K " + ~?}-

According to the theory of the stiffness of beams on end sup-
ports, deflections vary directly as the square of the length Z, and in-
versely as the distance c from the neutral axis to the remotest fibre
of the cross-section. Assuming that the deflections of columns

STRENGTH OF MATERIALS

follow the same laws, we may write d = ~k (I2 -=- c), where ~k is
some constant depending on the material of the column and on the
end conditions. Substituting this value for d in the last equation,
we find that

and

x =

P =

(10)

Each of these (usually the last) is known as " Rankme's formula."

For mild-steel columns a certain large steel company uses S = 50,000
pounds per square inch, and the following values of Jc:

1. Columns with two pin ends, k = 1 -r- 18,000.

2. " " one flat and one pin end, k = 1 -f- 24,000.

3. " " both ends flat, fc = 1 -*- 36,000.

With these values of S and fc, P of the formula means the ultimate load,
that is, the load causing failure. The safe load equals P divided by the
selected factor of safety— a factor of 4 for steady loads, and 5 for moving
loads, being recommended by the company referred to. The same unit is to
be used for Z and r.

Cast-iron columns are practically always made hollow with
comparatively thin walls, and are usually circular or rectangular
in cross -section. Tho following modifications of Rankine's formula
are sometimes used:

For circular sections, --r— =

For rectangular sections, -^ =

1,000 d2

(ID")

In these formulas d denotes the outside diameter of the circular sec-
tions or the length of the lesser side of the rectangular sections. The same
unit is to be used for I and d.

Examples. 1. A 40-pound 10-inch steel I-beam 8 feet
long is used as a flat-ended column. Its load being 100,000
pounds, what is its factor of safety ?

Obviously the column tends to bend in a plane perpendicular
to its web. Hence the radius of gyration to be used is the one

STRENGTH OF MATERIALS 91

with respect to that central axis of the cross-section which is in
the web, that is, axis 2-2 (see figure accompanying table, page 72) .
The moment of inertia of the section with respect to that axis,
according to the table, is 9.50 inches4; and since the area of the
section is 11.76 square inches,

9'5° --0808
"

"L76

Now, I = 8 feet = 96 inches; and since k = 1 -r- 36,000, and S =
50,000, the breaking load for this column, according to Rankine's
formula, is

p = 50,000 X 11.76 _

446,790 pounds.

1 +

36,000 X 0.808.

Since the factor of safety equals the ratio of the breaking load to
the actual load on the column, the factor of safety in this case is

446,790

2. What is the safe load for a cast-iron column 10 feet long
with square ends and a hollow rectangular section, the outside,
dimensions being 5x8 inches; the inner, 4x7 inches; and the
factor of safety, 6 ?

In this case I — 10 feet = 120 inches; A = 5 X8-4 X 7
= 12 square -inches; and d = 5 inches. Hence, according to
formula 10', for rectangular sections, the breaking load is

P = 80.000X^18 = 610>000 pound,

h 1,000 X 52

Since the safe load equals the breaking load divided by the factor
of safety, in this case the safe load equals

= 101,700 pounds.

3. A channel column (see Fig. 46, $) is pin-ended, the pins
being perpendicular to the webs of the channel (represented by
AA in the figure), and its length is 16 feet (distance between axes

92 STRENGTH OF MATERIALS

of the pins). If the sectional area is 23.5 square inches, and the
moment of inertia with respect to AA is 386 inches4 and with
respect to BB 214 inches4, what is the safe load with a factor of
safety of 4 ?

The column is liable to bend in one of two ways, namely, in
the plane perpendicular to the axes of the two pins, or in the plane
containing those axes.

(1) For bending in the first plane, the strength of the col-
umn is to be computed from the formula for a pin -ended column.
Hence, for this case, r2 = 386 -r- 23.5 = 16; and the breaking
load is

18,000 X 16
The safe load for this case equals — - — -^ - = 260,400 pounds.

A i

(2) If the supports of the pins are rigid, then the pins
stiffen the column as to bending in the plane of their axes, and the
strength of the column for bending in that plane should be com-
puted from the formula for the strength of columns with flat ends.
Hence, r* = 214 -f- 23.5 = 9.11, and thebreaking load is

P = -- ' nf^v iov = 1,056,000 pounds.

36,000 X 9.11
The safe load for this case equals - - = 264,000 pounds.

EXAMPLES FOR PRACTICE.

1. A 40-pound 12-inch steel I-beam 10 feet long is used as
a column with flat ends sustaining a load of 100,000 pounds.

What is its factor of safety?

Ans. 4.1

2. A cast-iron column 15 feet long sustains a load of
150,000 pounds. Its section being a hollow circle, 9 inches out-
side and 7 inches inside diameter, what is the factor of safety?

Ans. 8.9

3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and
has square ends; the least radius of gyration of its cross-section is

STRENGTH OF MATERIALS

3.1 inches; and the area of the cross-section is 24.5 square inches.
What is the safe load for the column with a factor of safety of 4 ?

Ans. 247,000 pounds.

4. A cast-iron column 13 feet long has a hollow circular
cross-section 7 inches outside and 5J inches inside diameter.
What is its safe load with a factor of safety of 6 ?

Ans. 121,142 pounds.

5. Compute the safe load for a 40-pound 12-inch steel
I-beam used as a column with flat ends, its length being 17 feet.
Use a factor of safety of 5.

Ans. 52,470 pounds.

84. Graphical Representation of Column Formulas. Col-
umn (and most other engineering) formulas can be represented
graphically. To represent Rankine's formula for flat-ended mild-
steel columns,

P 50,000

36,000

we first substitute different values of I -r- r in the formula, and
solve for P -r- A. Thus we find, when

Z - r = 40, P - A = 47,900 ;
I + r = 80, P - A = 42,500 ;
l + r = 120, P -*- A = 35,750 ;
etc., etc.

Now, if these values of I -s- r be laid off by some scale on a line
from O, Fig. 48, and the corresponding values of P -r- A be laid

,1-r-r

100

zoo

300

Fig. 48.

off vertically from the points on the line, we get a series of points
as #, &, 6', etc.; and a smooth curve through the points a, b, ct

94 STRENGTH OF MATERIALS

etc., represents the formula. Such a curve, besides representing
the formula to one's eye, can be used for finding the value of
P -5- A for any value of I -f- r; or the value of I -f- r for any value
of P -5- A. The use herein made is in explaining other column
formulas in succeeding articles.

85. Combination Column Formulas. Many columns have
been tested to destruction in order to discover in a practical way
the laws relating to the strength of columns of different kinds.
The results of such tests can be most satisfactorily represented
graphically by plotting a point in a diagram for each test. Thus,
suppose that a column whose I -r- r was 80 failed under a load of
276,000 pounds, and that the area of its cross-section was 7.12
square inches. This test would be represented by laying off Oa,
Fig. 49, equal to 80, according to some scale; and then ab equal to
276,000 -r- 7.12 (P H- A), according to some other convenient
scale. The point b would then represent the result of this par-
ticular test. All the dots in the figure represent the way in which
the results of a series of tests appear when plotted.

It will be observed at once that the dots do not fall upon any
one curve, as the curve of Rankine's formula. Straight lines and

50000 ••
AOOOO- «-
30OOO • •
2OOOO--

10OOO

,V*-r

1OO 2OO 300

Fig. 49.

curves simpler than the curve of Rankine's formula have been
fitted to represent the average positions of the dots as determined
by actual tests, and the formulas corresponding to such lines have
been deduced as column formulas. These are explained in the
following articles.

86. Straight-Line and Euler Formulas. It occurred to Mr.
T. H. Johnson that most of the dots corresponding to ordinary

STRENGTH OF MATERIALS

lengths of columns agree with a straight line just as well as with
a curve. He therefore, in 1886, made a number of such plats or
diagrams as Fig. 49, fitted straight lines to them, and deduced the
formula corresponding to each line. These have become known
as " straight-line formulas," and their general form is as follows:

— r- = S - m — ,

A T

(•I)

P, A, Z, and r having meanings as in Rankine's formula (Art. 83),
and S and m being constants whose values according to Johnson
are given in Table E below.

For the slender columns, another formula (Euler's, long since
deduced) was used by Johnson. Its general form is —

JP n

(12)

(I + rf

n being a constant whose values, according to Johnson, are given
in the following table:

TABLE E.

Data for Mild-Steel Columns.

Limit (I -i- r)

Hinged ends

52,500

220

160

444,000,000

Flat ends

52,500

180

195

666,000,000

The numbers in the fourth column of the table mark the point of divi-
sion between columns of ordinary length and slender columns. For the
former kind, the straight-line formula applies; and for the second, Euler's.
That is, if the ratio / •*• r for a steel column with hinged end, for example, is
less than 160, we must use the straight-line formula to compute its safe load,
factor of safety, etc.; but if the ratio is greater than 160, we must use Euler's
formula.

For cast-iron columns with flat ends, S = 34,000, and m = 88; and since
they should never be used " slender," there is no use of Euler's formula for
cast-iron columns.

The line AB, Fig. 50, represents Johnson's straight-line for-
mula; and BC, Euler's formula. It will be noticed that the two
lines are tangent; the point of tangency corresponds to the "lim-
iting value " I -r- r, as indicated in the table.

Examples. 1. A 40-pound 10-in^.h steel I-beam column 8

STRENGTH OF MATERIALS

feet long sustains a load of 100,000 pounds, and the ends are flat.
Compute its factor of safety according to the methods of this
article.

The first thing to do is to compute the ratio I -r- r for the
column, to ascertain whether the straight-line formula or Euler's

P-f-A

50000
4-OOOO-
30OOO
2OOOO-
JOOOO

100

200

300

Fig. 50.

formula should be used. From Table C, on page 70, we find that
the moment of inertia of the column about the neutral axis of
its cross -section is 9.50 inches4, and the area of the section is
11.76 square inches. Hence

9.50
1L76"

= 0.81; or r = 0.9 inch.

Since I = 8 feet = 96 inches,

J_ 96
r

0.9 - 106T

This value of I -5- r is less than the limiting value (195) indicated
by the table for steel columns with flat ends (Table E, p. 97), and
we should therefore use the straight-line formula; hence

= 52,500 - 180 X 106-jp

or, P = 11.76 (52,500 - 180 X 106-?-) = 391,600 pounds.

This is the breaking load for the column according to the straight,
line formula; hence the factor of safety is

391,600
100,000

STRENGTH OF MATERIALS 97

2. Suppose that the length of the column described in the
preceding example were 16 feet. What would its factor of safety be?

Since I = 16 feet = 192 inches; and, as before, r = 0.9
inch, I ~- r = 21 3 J. This value is greater than the limiting
value (195) indicated by Table E (p. 97) for flat-ended steel col-
umns; hence Euler's formula is to be used. Thus

P _ 666,000,000
IL76 " (213J)2 '

11.76 X 666,000,000
or, P = - « - ^ 172>100

This is the breaking load; hence the factor of safety is
172,100

100,000

1.7

3. What is the safe load for a cast-iron column 10 feet long
with square ends and hollow rectangular section, the outside
dimensions being 5x8 inches and the inside 4x7 inches, with a
factor of safety of 6 ?

Substituting in the formula for the radius of gyration given
in Table A, page 52, we get

r= , 8 X 53-7x 43

12 (8 X 5 - 7 X 4)
Since I = 10 feet = 120 inches,

1 12° 6122

~ T96 = 6L22

According to the straight-line formula for cast iron, A being
equal to 12 square inches,

= 34?000 - 88 X 61.22;

or, P = 12 (34,000 - 88 X 61.22) = 343,360 pounds.

This being the breaking load, the safe load is

= 57,227 pounds.

98 STRENGTH OF MATERIALS

EXAMPLES FOR PRACTICE.

1. A 40-pound 12-inch steel I-beam 10 feet long is used as
a flat-ended column. Its load being 100,000 pounds, compute
the factor of safety by the formulas of this article.

Ans. 3.5

2. A cast-iron column 15 feet long sustains a load of
150,000 pounds. Its section being a hollow circle of 9 inches
outside and 7 inches inside diameter, compute the factor of safety
by the straight-line formula.

Ans. 4.8

3. A steel Z-bar column (see Fig. 46, a] is 24 feet long
and has square ends; the least radius of gyration of its cross-
section is 3.1 inches; and the area of the cross-section is 24.5
square inches. Compute the safe load for the column by the
formulas of this article, using a factor of safety of 4.

Ans. 219,000 pounds.

4. A hollow cast-iron column 13 feet long has a circular
cross-section, and is 7 inches outside and 5J inches inside in
diameter. Compute its safe load by the formulas of this article,
using a factor of safety of 6.

Ans. 68,500 pounds

5. Compute by the methods of this article the safe load for
a 40-pound 12 -inch steel I-beam used as a column with flat ends,,
If the length is 17 feet and the factor of safety 5.

Ans. 35,100 pounds.

87. ParaboIa=EuIer Formulas. As better fitting the results
of tests of the strength of columns of " ordinary lengths," Prof.
J. B. Johnson proposed (1892) to use parabolas instead of straight
lines. The general form of the " parabola formula " is

P, A, I and r having the same meanings as in Rankine's formula,
Art. 83; and S and m denoting constants whose values, according
to Professor Johnson, are given in Table F below.

Like the straight-line formula, the parabola formula should
not be used for slender columns, but the following (Euler's) is
applicable:

STRENGTH OF MATERIALS

the values of n (Johnson) being given in the following table

TABLE F.

Data for Jlild Steel Columns.

Limit (I H- r)

Hinged ends

42,000

0.97

150

456,000,000

Flat ends . .

42.000

0.62

190

712,000,000

The point of division between columns of ordinary length and slender
columns is given in the fourth column of the table. That is, if the ratio Z-Hr
for a column with hinged ends, for example, is less than 150, the parabola
formula should be used to compute the safe load, factor of safety, etc.; but
if the ratio is greater than 150, then Euler's formula should be used.

The line AB, Fig. 51, represents the parabola formula ; and the line
BC, Euler's formula. The two lines are tangent, and the point of tangency
corresponds to the " limiting value" l-^-r of the table.

For wooden columns square in cross-section, it is convenient to replace
r by d, the latter denoting the length of the sides of the square. The formula
becomes

' :

S and m for flat-ended columns of various kinds of wood having the follow-
ing values according to Professor Johnson:

For White pine, 8=2,500, w = 0.6;

11 Short-leaf yellow pine, 8=3,300, m = 0.7j

" Long-leaf yellow pine, 8=4,000, m = 0.8;

11 White oak, 8=3,500, w = 0.8.

The preceding formula applies to any wooden column whose ratio, l-r-d,
is less than 60, within which limit columns of practice are included.

10000-

100 200

Fig. 51.

300

Examples. 1. A 40-pound 10-inch steel I-beam column

100 STRENGTH OF MATERIALS

8 feet long sustains a load of 100,000 pounds, and its ends are flat.
Compute its factor of safety according to the methods of this
article.

The first thing to do is to compute the ratio I -f- r for the
column, to ascertain whether the parabola formula or Euler's for-
mula should be used. As shown in example 1 of the preceding

article, I -s- r = 106§. This ratio being less than the limiting

-A .j
value, 190, of the table, we shoula use^he ;parabola formula.

Hence, since the area of the cross -"section- is 11.76 square inches
(see Table C, page 70), .-.«

-*-B = 42,000 -0.62 (106f)«;

or, P = 11.76 [42,000 - 0.62 (106f )2] = 410,970 pounds.

This is the breaking load according to the parabola formula; hence
the factor of safety is

410,970 __ 41

100,000 "

2. A white pine column 10 X 10 inches in cross-section and
18 feet long sustains a load of 40,000 pounds. What is its factor
of safety ?

The length is 18 feet or 216 inches; hence the ratio I -f- d =
21.6, and the parabola formula is to be applied.
Now, since A = 10 X 10 = 100 square inches,

j^- = 2,500 - 0.6 X 21.6';

or, P = 100 (2,500 - 0.6 X 21.62) = 222,000 pounds.

This being the breaking load according to the parabola formula,
the factor of safety is

222,000

40,000 =

3. What is the safe load for a long-leaf yellow pine column
12 X 12 inches square and 30 feet long, the factor of safety
being 5 ?

The length being 30 feet or 360 inches,

1 - - 36° so-

d - 12" ~ 3°>

STRENGTH OF MATERIALS , . , 101

hence the parabola formula should be used. Since A = 12 X 12
= 144 square inches,

JL = 4,000 - 0.8 X 302;

or, P == 144 (4,000 - 0.8 X 302) = 472,320 pounds.

This being the breaking load according to the parabola formula,
the safe load is

1J! - = 94,465 pounds.

EXAMPLES FOR PRACTICE.

1. A 40-pound 12-inch steel I-beam 10 feet long is used as
a flat-ended column. Its load being 100,000 pounds, compute its
factor of safety by the formulas of this article.

Ans. 3.8

2. A white oak column 15 feet long sustains a load of
30,000 pounds. Its section being 8x8 inches, compute the
factor of safety by the parabola formula.

Ans. 6.6

3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and
has square ends; the least radius of gyration of its cross-section
is 3.1 inches; and the area of its cross-section is 24.5 square
inches. Compute the safe load for the column by the formulas
of this article, using a factor of safety of 4.

Ans. 224,500 pounds.

4. A short-leaf yellow pine column 14 X 14 inches in sec-
tion is 20 feet long. What load can it sustain, with a factor of
safety of 6 ?

Ans. 101,100 pounds.

88. " Broken Straight-Line " Formula. A large steel com-
pany computes the strength of its flat-ended steel columns by two
formulas represented by two straight lines AB and BC, Fig. 52.
The formulas are

X = 48,000,

and -?- = 68,400 - 228 £,

P. A, I, and r having the same meanings as in Art. 83,

102

STRENGTH OF MATERIALS

The point B corresponds very nearly to the ratio I ~- r = 90.
Hence, for columns for which the ratio Z -r- r is less than 90, the
first formula applies; and for columns for which the ratio is
greater than 90, the second one applies. The point C corre-
sponds to the ratio Z -r- r = 200, and the second formula does not
apply to a column for which I -f- r is greater than that limit.

P-J-A

50000

10OOO

The ratio I -r- r for steel columns of practice rarely Exceeds 150,
and is usually less than 100.

Fig. 53 is a combination of Figs. 49, 50, 51 and 52, and
represents graphically a comparison of the Rankine, straight-line,
Euler, parabola -Euler, and broken straight-line formulas for flat-
ended niild-steel columns, It well illustrates the fact that our
knowledge of the strength of columns is not so exact as that, for
example, of the strength of beams.

100

200

3OO

Fig. 53.

89. Design of Columns. All the preceding examples relat-
ing to columns were on either (1) computing the factor of safety

STRENGTH OF MATERIALS 103

of a given loaded column, or (2) computing the safe load for a
given column. A more important problem is to design a column
to sustain a given load under given conditions. A complete dis-
cussion of this problem is given in a later paper on design. "We
show here merely how to compute the dimensions of the cross -
section of the column after the form of the cross-section has been
decided upon.

In only a few cases can the dimensions be computed directly
(see example 1 following), but usually, when a column formula is
applied to a certain case, there will be two unknown quantities in
it, A and r or d. Such cases can best be solved by trial (see
examples 2 and 3 below).

Example. 1. What is the proper size of white pine column
to sustain a load of 80,000 pounds with a factor of safety of 5,
when the length of the column is 22 feet ?

We use the parabola formula (equation 13). Since the safe
load is 80,000 pounds and the factor of safety is 5, the breaking
load P is

80,000 X 5 = 400,000 pounds.

The unknown side of the (square) cross-section being denoted by
d, the area A is d2. Hence, substituting in the formula, since I
= 22 feet = 264 inches, we have

=2,500- 0.6

Multiplying both sides by d2 gives

400,000 = 2,500 d2 - 0.6 X 2642,

or 2,500 d* = 400,000 + 0.6 x 264* = 441,817.6.

Hence d2 = 176.73, or d = 13.3 inches.

2. What size of cast-iron column is needed to sustain a load
of 100,000 pounds with a factor of safety of 10, the length of the
column being 14 feet ?

We shall suppose that it has been decided to make the cross-
section circular, and shall compute by Rankine's formula modified
for cast-iron columns (equation 10'). The breaking load for the
column would be

104 STRENGTH OF MATERIALS

100,000 X 10 = 1,000,000 pounds.

The length is 14 feet or 168 inches; hence the formula oecomes
1,000,000 80,000

or, reducing by dividing both sides of the equation by 10,000, and
then clearing of fractions, we have

100

There are two unknown quantities in this equation, d and A, and
we cannot solve directly for them. Probably the best way to pro-
ceed is to assume or guess at a practical value of d, then solve for
A, and finally compute the thickness or inner diameter. Thus, let
us try d equal to 7 inches, first solving the equation for A as far
as possible. Dividing both sides by 8 we have

100 n 1682

A . _ i I

and, combining,

441
A = 12.5 + ~.

Now, substituting 7 for d, we have

441
A = 12.5 + -JQ- = 21.5 square inches.

The area of a hollow circle whose outer and inner diameters are
d and dl respectively, is 0.7854 (d2 - d*). Hence, to find the inner
diameter of the column, we substitute 7 for d in the last expres-
sion, equate it to the value of A just found, and solve for dt. Thus,

0.7854 (49-^) = 21.5-
hence

and d* = 49 - 27.37 = 21.63 or dl = 4.65.

This value of d makes the thickness equal to

J (7-4.65) = 1.175 inches,

STRENGTH OF MATERIALS 105

which is safe. It might be advisable in an actual case to try
d equal to 8 repeating the computation.*

EXAMPLE FOR PRACTICE.

1. What size of white oak column is needed to sustain a load
of 45,000 pounds with a factor of safety of 6, the length of the
column being 12 feet.

Ans. d = 8-J, practically a 10 X 10-inch section

STRENGTH OF SHAFTS.

A shaft is a part of a machine or system of machines, and is
used to transmit power by virtue of its torsional strength, or resist-
ance to twisting. Shafts are almost always made of metal and are
usually circular in cross-section, being sometimes made hollow.

90. Twisting Moment. Let AF, Fig. 54, represent a shaft
with four pulleys on it. Suppose that D is the driving pulley
and that B, C and E are pulleys from which power is taken off to
drive machines. The portions of the shafts between the pulleys

Pig. 54.

are twisted when it is transmitting power; and by the twisting
moment at any cross-section of the shaft is meant the algebraic
sum of the moments of all the forces acting on the shaft on either

*NOTE. The structural steel handbooks contain extensive tables by
means of which the design of columns of steel or cast iron is much facilitated.
The difficulties encountered in the use of formulae are well illustrated in this
example.

106 STRENGTH OF MATERIALS

side of the section, the moments being taken with respect to the
axis of the shaft. Thus, if the forces acting on the shaft (at the
pulleys) are P15 P2, P3, and P4 as shown, and if the arms of the
forces or radii of the pulleys are al9 az, #3, and a^ respectively, then
the twisting moment at any section in

BC is Pj «„

CD is P, a, + P2 «a,

DE is Pj a, + P2 a2 - P3 a3.

Like bending moments, twisting moments are usually ex-
pressed in inch -pounds.

Example. Let at = a2 = a4 = 15 inches, a3 == 30 inches,
P, = 400 pounds, P2 = 500 pounds, P3 = 750 pounds, and P4 =
600 pounds.* What is the value of the greatest twisting moment
in the shaft ?

At any section between the first and second pulleys, the
twisting moment is

400 X 15 = 6,000 inch-pounds;

at any section between the second and third it is

400 X 15 + 500 X 15 = 13,500 inch-pounds; and
at any section between the third and fourth it is

400 X 15 + 500 X 15 - 750 X 30 = - 9,000 inch-pounds.
Hence the greatest value is 13,500 inch-pounds.

91. Torsional Stress. The stresses in a twisted shaft are
called "torsional" stresses. The torsional stress on a cross -section
of a shaft is a shearing stress, as in the case illustrated by Fig. 55,
which represents a flange coupling in a shaft. Were it not for
the bolts, one flange would slip over the other when either part
of the shaft is turned; but the bolts prevent the slipping. Obvi-
ously there is a tendency to shear the bolts off unless they are
screwed up very tight; that is, the material of the bolts is sub-
jected to shearing stress.

Just so, at any section of the solid shaft there is a tendency
for one part to slip past the other, and to prevent the slipping or

* Note. These numbers were so chosen that the moment of P (driving
moment) equals the sum of the moments of the other forces. This is always
the case in a shaft rotating at constant speed; that is, the power given the
shaft equals the power taken off.

STRENGTH OF MATERIALS

107

shearing of the shaft, there arise shearing stresses at all parts of
the cross -section. The shearing stress on the cross-section of a
shaft is not a uniform stress, its value per unit-area being zero at
the center of the section, and increasing toward the circumference.
In circular sections, solid or hollow, the shearing stress per unit-
area (unit-stress) varies directly as the distance from the center
of the section, provided the elastic limit is not exceeded. Thus,
if the shearing unit-stress at the circumference of a section is

Fig. 55.

1,000 pounds per square inch, and the diameter of the shaft is
2 inches, then, at -J inch from the center, the unit-stress is 500
pounds per square inch; and at J inch from the center it is 250
pounds per square inch. In Fig. 55 the arrows indicate the
values and the directions of the shearing stresses on very small
portions of the cross-section of a shaft there represented.

92. Resisting Moment. By "resisting moment" at a sec-
tion of a shaft is meant the sum of the moments of the shearing
stresses on the cross- section about the axis of the shaft.

Let Ss denote the value of the shearing stress per unit-area
(unit-stress) at the outer points of a section of a shaft; d the
diameter of the section (outside diameter if the shaft is hollow);
and dl the inside diameter. Then it can be shown that the re-
sisting moment is:

For a solid section, 0.1963 Ss d3;

0.1963 Ss (tf - df)

lor a hollow section, *

93. Formula for the Strength of a Shaft. As in the case

108 STRENGTH OF MATERIALS

of beams, the resisting moment equals the twisting moment at
any section. If T be used to denote twisting moment, then we
have the formulas :

For solid circular shafts, 0.1963 Ss d? = T;

w VTI • i - -u ** 0.1963 Ss (d* - d*) r
For hollow circular shafts, - J-^ - L^ =

d }

In any portion of a shaft of constant diameter, the unit-
shearing stress Ss is greatest where the twisting moment is greatest.
Hence, to compute the greatest unit-shearing stress in a shaft,
we first determine the value of the greatest twisting moment,
substitute its value in the first or second equation above, as the
case may be, and solve for Ss. It is customary to express T in
inch-pounds and the diameter in inches, Ss then being in pounds
per square inch.

Examples. 1. Compute the value of the greatest shearing
unit-stress in the portion of the shaft between the first and second
pulleys represented in Fig. 54, assuming values of the forces and
pulley radii as given in the example of article 90. Suppose also
that the shaft is solid, its diameter being 2 inches.

The twisting moment T at any section of the portion between
the first and second pulleys is 6,000 inch-pounds, as shown in the
example referred to. Hence, substituting in the first of the two
formulas 15 above, we have

0.1963 Ss X 23 == 6,000;

fi 000

or, Ss = n 1Q' xx o = 3,820 pounds per square inch.

/\ o

This is the value of the unit-stress at the outside portions of all
sections between the first and second pulleys.

2. A hollow shaft is circular in cross-section, and its outer
and inner diameters are 16 and 8 inches respectively. If the
working strength of the material in shear is 10,000 pounds per
square inch, what twisting moment can the shaft safely sustain ?

The problem requires that we merely substitute the values of
Ss, <?, and d^ in the second of the above formulas 15, and solve for
T. Thus,

rp 0.1963 X 10,000 (16* - 84) - ™ Q0n . ,

- - JL_ — — L = 7,537,920 inch-pounds.

STRENGTH OF MATERIALS 109

EXAMPLES FOR PRACTICE.

1. Compute the greatest value of the shearing unit-stress in
the shaft represented in Fig. 54, using the values of the forces
and pulley radii given in the example of article 90, the diameter
of the shaft being 2 inches.

Ans. 8,595 pounds per square inch

2. A solid shaft is circular in cross-section and is 9.6 inches
in diameter. If the working strength of the material in shear is
10,000 pounds per square inch, how large a twisting moment can
the shaft safely sustain? (The area of the cross -section is practically
the same as that of the hollow shaft of example 2 preceding.)

Ans. 1,736,736 inch-pounds.

94. Formula for the Power Which a Shaft Can Transmit.
The power that a shaft can safely transmit depends on the shear-
ing working strength of the material of the shaft, on the size of
the cross-section, and on the speed at which the shaft rotates.

Let H denote the amount of horse -power; Ss the shearing
working strength in pounds per square inch; d the diameter
(outside diameter if the shaft is hollow) in inches; dl the inside
diameter in inches if the shaft is hollow; and n the number of
revolutions of the shaft per minute. Then the relation between
power transmitted, unit-stress, etc., is:

For solid shafts, H = 321 QOQ '*'

L (16)

O '

For hollow shafts, H =

. 1. What horse-power can a hollow shaft 16
inches and 8 inches in diameter safely transmit at 50 revolutions
per minute, if the shearing working strength of the material is
10,000 pounds per square inch?

We have merely to substitute in the second of the two for-
mulas 16 above, and reduce. Thus,

II = 321 QOQ X 16 -- = 6j0°° horse-Power (nearlj)-

2. What size of solid shaft is needed to transmit 6,000 horse-
power at 50 revolutions per minute if the shearing working
strength of the material is 10,000 pounds per square inch?

110 STRENGTH OF MATERIALS

We have merely to substitute in the first of the two formulas
16, and solve for d. Thus,

10,000 X <P X 50
6'0( 321,000 "'

6,000 X 321,000
therefore df 1U)00o x 50 = 3>852;

or, d = 3852 = 15.68 inches.

(A solid shaft of this diameter contains over 25% more material than
the hollow shaft of example 1 preceding. There is therefore considerable
saving of material in the hollow shaft.)

3. A solid shaft 4 inches in diameter transmits 200 horse-
power while rotating at 200 revolutions per minute. "What is the
greatest shearing unit-stress in the shaft?

We have merely to substitute in the first of the equations 16,
and solve for Ss. Thus,

X 43 X 200

200 =

321,000

200 X 321,000
or, S = 43 \/ 200 = ">. " pounds per square inch.

EXAMPLES FOR PRACTICE.

1. What horse-power can a solid shaft 9.6 inches in diameter
safely transmit at 50 revolutions per minute, if its shearing work-
ing strength is 10,000 pounds per square inch ?

Ans. 1,378 horse-power.

2. What size of solid shaft is required to transmit 500 horse-
power at 150 revolutions per minute, the shearing working strength
of the material being 8,000 pounds per square inch.

Ans. 5.1 inches.

3. A hollow shaft whose outer diameter is 14 and inner 6.7
inches transmits 5,000 horse-power at 60 revolutions per minute.
What is the value of the greatest shearing unit-stress in the shaft?

Ans. 10,273 pounds per square inch.

STIFFNESS OF RODS, BEAMS, AND SHAFTS.

The preceding discussions have related to the strength of

STRENGTH OF MATERIALS 111

materials. We shall now consider principally the elongation of
rods, deflection of beams, and twist of shafts.

95. Coefficient of Elasticity. According to Hooke's Law
(Art. 9, p. 7), the elongations of a rod subjected to an increasing
pull are proportional to the pull, provided that the stresses due to
the pull do not exceed the elastic limit of the material. Within
the elastic limit, then, the ratio of the pull and the elongation is
constant; hence the ratio of the unit-stress (due to the pull) to the
unit-elongation is also constant. This last-named ratio is called
" coefficient of elasticity." If E denotes this coefficient, S the
unit-stress, and s the unit-deformation, then

«-T> 07)

Coefficients of elasticity are usually expressed in pounds per square inch.

The preceding remarks, definition, and formula apply also to
a case of compression, provided that the material being compressed
does not bend, but simply shortens in the direction of the com-
pressing forces. The following table gives the average values of
the coefficient of elasticity for various materials of construction:

TABLE Q.

Coefficients of Elasticity.

Material.

Average Coefficient of Elasticity.

Steel

30,000,000 pounds per square inch.

Wrought iron
Cast iron

27,500,000
15,000,000 " " " "

Timber .

1,800,000 " " " "

The coefficients of elasticity for steel and wrought iron, for different
grades of those materials, are remarkably constant; but for different grades
of cast iron the coefficients range from about 10,000,000 to 30,000,000 pounds
per square inch. Naturally the coefficient has not the same value for the
different kinds of wood; for the principal woods it ranges from 1,600,000
(for spruce) to 2,100,000 (for white oak).

Formula 17 can be put in a form more convenient for use, as
follows :

Let P denote the force producing the deformation ; A the
area of the cross -section of the piece on which P acts ; I the length
of the piece ; and D the deformation (elongation or shortening).

11? STRENGTH OF MATERIALS

Then

S = P -j- A (see equation 1),
and s = D -+- I (see equation 2).

Hence, substituting these values in equation 17, we have

d7')

The first of these two equations is used for computing the value of
the coefficient of elasticity from measurements of a " test," and
the second for computing the elongation or shortening of a given
rod or bar for which the coefficient is known.

Examples. 1. It is required to compute the coefficient of
elasticity of the material the record of a test of which is given on
page 9.

Since the unit-stress S and unit-elongation s are already
computed in that table, we can use equation 17 instead of the first
of equations 17'. The elastic limit being between 40,000 and
45,000 pounds per square inch, we may use any value of the
unit-stress less than that, and the corresponding unit-elongation.

Thus, with the first values given,

5,000

With the second,

This lack of constancy in the value of E as computed from different
loads in a test of a given material, is in part due to errors in measuring the
deformation, a measurement difficult to make. The value of the coefficient
adopted from such a test, is the average of all the values of E which can be
computed from the record.

2. How much will a pull of 5,000 pounds stretch a round
steel rod 10 feet long and 1 inch in diameter ?

We use the second of the two formulas 17'. Since A =
0.7854 X I2 == 0.7854 square inches, I ?* 120 inches, and E -
30,000,000 pounds per square inch, the stretch is:

STRENGTH OF MATERIALS 113

EXAMPLES FOR PRACTICE.

1. "What is the coefficient of elasticity of a material if a pull
of 20,000 pounds will stretch a rod 1 inch in diameter and 4 feet

long 0.045 inch ?

Ans. 27,000,000 pounds per square inch.

2. How much will a pull of 15,000 pounds elongate a round
cast-iron rod 10 feet long and 1 inch in diameter ?

Ans. 0.152 inch.

96. Temperature Stresses. In the case of most materials,
when a bar or rod is heated, it lengthens; and \vhen cooled, it
shortens if it is free to do so. The coefficient of linear expansion
of a material is the ratio which the elongation caused in a rod or
bar of the material by a change of one degree in temperature bears
to the length of the rod or bar. Its values for Fahrenheit degrees
are about as follows:

For Steel, 0.0000065.

For Wrought iron, .0000067.
For Cast iron, .0000062.

Let K be used to denote this coefficient; t a change of tem-
perature, in degrees Fahrenheit; I the length of a rod or bar;
and D the change in length due to the change of temperature.
Then

D = = K tl. (18)

D and Z are expressed in the same unit.

If a rod or bar is confined or restrained so that it cannot
change its length when it is heated or cooled, then any change in
its temperature produces a stress in the rod; such are called tem-
perature stresses.

Examples. 1. A steel rod connects two solid walls and is
screwed up so that the unit-stress in it is 10,000 pounds per
square inch. Its temperature falls 10 degrees, and it is observed
that the walls have not been drawn together. What is the temper-
ature stress produced by the change of temperature, and what is
the actual unit-stress in the rod at the new temperature ?

Let I denote the length of the rod. Then the change in
length which would occur if the rod were free, is given by formula
18, above, thus:

D = 0.0000065 X 10 X I = 0.000065 I.

114 STRENGTH OF MATERIALS

Now, since the rod could not shorten, it has a greater than normal
length at the new temperature; that is, the fall in temperature has
produced an effect equivalent to an elongation in the rod amount-
ing to D, and hence a tensile stress. This tensile stress can be
computed from the elongation D by means of formula 17. Thus,

S = Es;
and since s, the unit-elongation, equals

D = .0000065 I = JMOO(ft

i i

S = 30,000,000 X .0000065 = 195.0 pounds per square inch.
This is the value of the temperature stress; and the new unit-
stress equals

10,000 + 195.0 = 10,195 pounds per square inch.

Notice that the unit temperature stresses are independent of the length
of the rod and the area of its cross-section.

2. Suppose that the change of temperature in the preceding
example is a rise instead of a fall. What are the values of the
temperature stress due to the change, and of the new unit- stress in
the rod ?

The temperature stress is the same as in example 1, that is,
1,950 pounds per square inch ; but the rise in temperature
releases, as it were, the stress in the rod due to its being screwed
up, and the final unit stress is

10,000 - 1,950 = 8,050 pounds per square inch.

EXAHPLE FOR PRACTICE,

1. The ends of a wrought-iron rod 1 inch in diameter are

fastened to two heavy bodies which are to be drawn together, the
j &

temperature of the rod being 200 degrees when fastened to the ob-
jects. A fall of 120 degrees is observed not to move them.
What is the temperature stress, and what is the pull exerted by
the rod on each object ?

( Temperature stress, 22,000 pounds per square inch.
AnS' \ Pull, 17,280 pounds.

97. Deflection of Beams. Sometimes it is desirable to know
how much a given beam will deflect under a given load, or to design

STRENGTH OF MATERIALS 115

a beam which will not deflect more than a certain amount under a
given load. In Table B, page 53, Part I, are given formulas for
deflection in certain cases of beams and different kinds of loading.

In those formulas, d denotes deflection; I the moment of inertia of the
cross-section of the beam with respect to the neutral axis, as in equation 6 ;
and E the coefficient of elasticity of the material of the beam (for values, see
Art. 95).

In each case, the load should be expressed in pounds, the length in
inches, and the moment of inertia in biquadratic inches; then the deflection
will be in inches.

According to the formulas for d, the deflection of a beam
varies inversely as the coefficient of its material (E) and the mo-
ment of inertia of its cross-section (I) ; also, in the first four and
last two cases of the table, the deflection varies directly as the cube
of the length (Z3).

Example. What deflection is caused by a uniform load of
6,400 pounds (including weight of the beam) in a wooden beam
on end supports, which is 12 feet long and 6 X 12 inches in
cross -section ? (This is the safe load for the beam ; see example
1, Art. 65.)

The formula for this case (see Table B, page 53) is

5 W
= 384 El '

Here W = 6,400 pounds ; I = 144 inches ; E = 1,800,000
pounds per square inch ; and

I = -^ la? = jg- 6 X 123= 864 inches4.

Hence the deflection is

5 X 6,400 X 144*
= 384 X 1,800,000 X 864 =

EXAMPLES FOR PRACTICE.

1. Compute the deflection of a timber built-in cantilever
8X8 inches which projects 8 feet from the wall and bears an
end load of 900 pounds. (This is the safe load for the cantilever,
see example 1, Art. 65.)

Ans. 0.43 inch.

2. Compute the deflection caused by a uniform load of 40,000

116

STRENGTH OF MATERIALS

pounds on a 42-pound 15-inch steel I-beam which is 16 feet long
and rests on end supports.

Ans. 0.28 inch.

98. Twist of Shafts. Let Fig. 57 represent a portion of a
shaft, and suppose that the part represented lies wholly between

Fig. 57.

two adjacent pulleys on a shaft to which twisting forces are applied
(see Fig. 54). Imagine two radii ma and nb in the ends of the
portion, they being parallel as shown when the shaft is not twisted.
After the shaft is twisted they will not be parallel, ma having
moved to ma', and nb to nb1 '. The angle between the two lines in
their twisted positions (ma' and nb') is called the angle of twist,
or angle of torsion, for the length 1. If a a" is parallel to ab, then
the angle a"nb' equals the angle of torsion.

If the stresses in the portion of the shaft considered do not
exceed the elastic limit, and if the twisting moment is the same
for all sections of the portion, then the angle of torsion a (in
degrees) can be computed from the following:

For solid circular shafts,

a =

584 TZ 36,800,000 HI

For hollow circular shafts,

(19)

584 Tld 36,800,000 HZ

Here T, Z, d, dl9 H, and n have the same meanings as in Arts. 93
and 94, and should be expressed in the units there used. The
letter E1 stands for a quantity called coefficient of elasticity for
shear; it is analogous to the coefficient of elasticity for tension and
compression (E), Art. 95. The values of E1 for a few materials
average about as follows (roughly E1 = | E) :

STRENGTH OF MATERIALS 117

For Steel, 11,000,000 pounds per square inch.

For Wrought iron, 10,000,000 " " " "
For Cast iron, 6,000,000 " " « "

Example. What is the value of the angle of torsion of a
steel shaft 60 feet long when transmitting 6,000 horse-power at
50 revolutions per minute, if the shaft is hollow and its outer and
inner diameters are 16 and 8 inches respectively ?

Here I — 720 inches; hence, substituting in the appropriate
formula (19), we find that

36,800,000 X 6,000 x
= 11,000,000 X (16- - 80 50 =

EXAMPLE FOR PRACTICE.

Suppose that the first two pulleys in Fig. 54 are 12 feet
apart; that the diameter of the shaft is 2 inches; and that P, = 400
pounds, and al = 15 inches. If the shaft is of wrought iron,
what is the value of the angle of torsion for the portion between
the first two pulleys ?

Ans. 3.15 degrees.

99. Non-elastic Deformation. The preceding formulas for
elongation, deflection, and twist hold only so long as the greatest
unit-stress does not exceed the elastic limit. There is no theory,
and no formula, for non -elastic deformations, those corresponding
to stresses which exceed the elastic limit. It is well known, how-
ever, that non -elastic deformations are not proportional to the
forces producing them, but increase much faster than the loads.
The value of the ultimate elongation of a rod or bar (that is, the
amount of elongation at rupture), is quite well known for many
materials. This elongation, for eight-inch specimens of various
materials (see Art. 16), is :

For Cast iron, about 1 per cent.
For Wrought iron (plates), 12 - 15 per cent.
For " " (bars), 20-25 " " .

For Structural steel, 22-26 " " .

Specimens of ductile materials (such as wrought iron and
structural steel), when pulled to destruction, neck down, that is,
diminish very considerably in cross-section at some place along
the length of the specimen. The decrease in cross-sectional area

118

STRENGTH OF MATERIALS

is known as reduction of area, and its value for wrought iron and
steel may be as much as 50 per cent.

RIVETED JOINTS.

100. Kinds of Joints. A lap joint is one in which the
plates or bars joined overlap each other, as in Fig. 58, a. A butt
joint is one in which the plates or bars that are joined butt against
each other, as in Fig. 58, b. The thin side plates on butt joints

Fig. 58.

are called cover=plates ; the thickness of each is always made not
less than one-half the thickness of the main plates, that is, the
plates or bars that are joined. Sometimes butt joints are made
with only one cover-plate; in such a case the thickness of the
cover-plate is made not less than that of the main plate.

"When wide bars or plates are riveted together, the rivets are
placed in rows, always parallel to the " seam " and sometimes also
perpendicular to the seam; but when we speak of a row of rivets,
we mean a row parallel to the seam. A lap joint with a single
row of rivets is said to be single=riveted ; and one with two rows
of rivets is said to be double-riveted. A butt joint with two rowa
of rivets (one on each side of the joint) is called " single-riveted,"
and one with four rows (two on each side) is said to be "double-
riveted."

The distance between the centers of consecutive holes in a
row of rivets is called pitch.

101. Shearing Strength, or Shearing Value, of a Rivet.
When a lap joint ia subjected to tension (that is, when P, Fig. 58,
#, is a pull), and when the joint is subjected to compression (when
P is a push), there is a tendency to cut or shear each rivet along
the surface between the two plates. In butt joints with two cover-

STRENGTH OF MATERIALS 119

plates, there is a tendency to cut or shear each rivet on two sur-
faces (see Fig. 58, b). Therefore the rivets in the lap joint are
said to be in single shear ; and those in the butt joint (two covers)
are said to be in double shear.

The " shearing value " of a rivet means the resistance which
it can safely offer to forces tending to shear it on its cross -section.
This value depends on the area of the cross-section and on the work-
ing strength of the material. Let d denote the diameter of the
cross-section, and Ss the shearing working strength. Then, since
the area of the cross -section equals 0.7854 d2, the shearing strength
of one rivet is :

For single shear, 0.7854 d2 S8 .

For double shear, 1.5708 <& S, .

102. Bearing Strength, or Bearing Value, of a Plate. When
a joint is subjected to tension or compression, each rivet presses
against a part of the sides of the holes through which it passes.
By " bearing value " of a plate (in this connection) is meant the
pressure, exerted by a rivet against the side of a hole in the plate,
which the plate can safely stand. This value depends on the
thickness of the plate, on the diameter of the rivet, and on the
compressive working strength of the plate. Exactly how it
depends on these three qualities is not known; but the bearing
value is always computed from the expression t d Sc, wherein t
denotes the thickness of the plate; 6?, the diameter of the rivet or
hole; and Sc, the working strength of the plate.

103. Frictional Strength of a Joint. When a joint is sub-
jected to tension or compression, there is a tendency to slippage
between the faces of the plates of the joint. This tendency is
overcome wholly or in part by frictional resistance between the
plates. The frictional resistance in a well-made joint may be
very large, for rivets are put into a joint hot, and are headed or
capped before being cooled. In cooling they contract, drawing the
plates of the joint tightly against each other, and producing a
great pressure between them, which gives the joint a correspond-
ingly large frictional strength. It is the opinion of some that
all well-made joints perform their service by means of their
frictional strength; that is to say, the rivets act only by pressing
the plates together and are not under shearing stress, nor

120 STRENGTH OF MATERIALS

are the plates under compression at the sides of their holes. The
" frictional strength " of a joint, however, is usually regarded as
uncertain, and generally no allowance is made for friction in com-
putations on the strength of riveted joints.

104. Tensile and Compressive Strength of Riveted Plates.
The holes punched or drilled in a plate or bar weaken its tensile
strength, and to compute that strength it is necessary to allow for
the holes. By net section, in this connection, is meant the small-
est cross-section of the plate or bar ; this is always a section along
a line of rivet holes.

If, as in the foregoing article, t denotes the thickness of the
plates joined ; d, the diameter of the holes; nl9 the number of riv-
ets in a row ; and w, the width of the plate or bar ; then the net
section = (w - ntd) t.

Let St denote the tensile working strength of the plate ; then
the strength of the unriveted plate is wtSi9 and the reduced tensile
strength is (w - n^) t St.

The compressive strength of a plate is also lessened by the
presence of holes ; but when they are again filled up, as in a joint,
the metal is replaced, as it were, and the compressive strength of
the plate is restored. No allowance is therefore made for holes in
figuring the compressive strength of a plate.

105. Computation of the Strength of a Joint. The strength
of a joint is determined by either (1) the shearing value of the
rivets ; (2) the bearing value of the plate ; or (3) the tensile
strength of the riveted plate if the joint is in tension. Let P8 de-
note the strength of the joint as computed from the shearing
values of the rivets ; Pc, that computed from the bearing value of
the plates ; and Pt, the tensile strength of the riveted plates.
Then, as before explained,

Pt= (W - n,d) *St; J

P8= n2 0.7854 ^2S8; and V (20)

nz denoting the total number of rivets in the joint ; and na denot-
ing the total number of rivets in a lap joint, and one-half the
number of rivets in a butt joint.

Examples. 1. Two half-inch plates 7^ inches wide are con-

STRENGTH OF MATERIALS 121

nected by a single lap joint double-riveted, six rivets in two rows.
If the diameter of the rivets is | inch, and the working strengths
are as follows : St= 12,000, S§== 7,500, and Sc= 15,000 pounds
per square inch, what is the safe tension which the joint can
transmit ?

Here nl== 3, n= 6, and n3= 6 ; hence

Pt= (7-i- - 3 X ~) X -i- X 12,000 == 31,500 pounds;

Ps= 6 X 3.7854 X (-|-)2 X 7,500 = 19,880 pounds ;

PA*= 6 X -5- X--J- X 15,000 = 33,750 pounds.

Since P8 is the least of these three values, the strength of the
joint depends on the shearing value of its rivets, and it equals
19,880 pounds.

2. Suppose that the plates described in the preceding example
are joined by means of a butt joint (two cover-plates), and 12
rivets are used, being spaced as before. What is the safe tension
which the joint can bear ?

Here nl == 3, n2 = 12, and n3 = 6; hence, as in the preced-
ing example,

Pt = 31,500; and Pc = 33,750 pounds; but

Ps = 12 X 0.7854 X (-|-)2 X 7,500 = S9,760 pounda

The strength equals 31,500 pounds, and the joint is stronger than
the first.

3. Suppose that in the preceding example the rivets are
arranged in rows of two. What is the tensile strength of the
joint ?

Here nl = 2. n2 — 12, and n3 == 6; hence, as in the preced-
ing example,

Ps == 39,760; and Pc = 33,750 pounds; but

Pt = (7 -|--2 X -|-) ^- X 12,000 = 36,000 pounds.

The strength equals 33,750 pounds, and this joint is stronger than
either of the first two.

122 STRENGTH OF MATERIALS

EXAMPLES FOR PRACTICE.

Note. Use working strengths as in example 1, above.
St = 12,000, S8 = 7,500, and Sc = 15,000 pounds per square inch.

1. Two half-inch plates 5 inches wide are connected by a
lap joint, with two |-inch rivets in a row. "What is the safe
strength of the joint ?

Ans. 6,625 pounds.

2. Solve the preceding example supposing that four |-inch
rivets are used, in two rows.

Ans. 13,250 pounds.

3. Solve example 1 supposing that three 1-inch rivets are
used, placed in a row lengthwise of the joint.

Ans. 17,670 pounds.

4. Two half -inch plates 5 inches wide are connected by a
butt joint (two cover-plates), and four |-inch rivets are used, in
two rows. What is the strength of the joint ?

Ans. 11,250 pounds.

106. Efficiency of a Joint. The ratio of the strength of a
joint to that of the solid plate is called, the " efficiency of the
joint." If ultimate strengths are used in computing the ratio,
then the efficiency is called ultimate efficiency; and if working
strengths are used, then it is called working efficiency. In the
following, we refer to the latter. An efficiency is sometimes ex-
pressed as a per cent. To express it thus, multiply the ratio
strength of joint --f- strength of solid plate, by 100.

Example. It is required to compute the efficiencies of the
joints described in the examples worked out in the preceding article.

In each case the plate is -| inch thick and 7J inches wide;
hence the tensile working strength of the solid plate is

7-L x _L x 12,000 = 45,000 pounds.
Therefore the efficiencies of the joints are :

(2) = 0.70, or 70 per cent;

INDEX

Page

Angle of torsion 116

Angle of twist 116

Applications of first beam formula 59

Beam formula

first 59

second 73

Beams 19

deflection of 114

stiffness of 110

Bearing strength of a plate. . 119

Bending moment 23, 32

maximum 40

Brick, ultimate strength of 15

Broken straight-line formula 101

Built-up sections, moment of inertia of 49

Butt joint 118

Cast iron

in tension v 13

in compression 15

Center of gravity of an area , 41

Center of gravity of built-up sections 44

Centrally loaded column 88

Coefficient of elasticity Ill

for shear 116

Coefficient of linear expansion 113

Column formulas

graphical representation of 93

Rankine's 88

Column loads, kinds of 88

Columns

classes of 86

cross-sections of ; 86

design of 102

end conditions of 85

strength of 85

Combination column formulas 94

Combined flexural and direct stress 83

Compression, materials in 13

Compressive strength of riveted plates 120

Continuous beam . . 19

124 INDEX

Page

Cover-plates 118

Deflection of beams 114

Deformation 4

Design of

columns 102

timber beams 76

Direct stress 79

Double shear 119

Eccentrically loaded ; 88

Efficiency of a joint 122

Elastic limit 5

Elasticity , 4

End conditions of column 85

Euler formulas 94

External shear 23

Factor of safety 8

Fibre stresses 55, 57

First beam formula • 59

Flexural stress 79

Flexure and compression 81

Flexure and tension 79

Formula for power which shaft can transmit 109

Formula for strength of shaft 107

Frictional strength of joint 119

Graphical representation of column formulas 93

Hooke's law 5

Horizontal shear 75

Inclined forces 54

Joint

efficiency of 122

frictional strength of 119

Kinds of loads and beams 52, 79

Kinds of stress 2, 55

Lap joint 118

Laws of strength of beams 71

Linear expansion, coefficient of 113

Longitudinal forces 54

Main plates 118

Materials

in compression 13

in shear 15

in tension 11

Maximum bending moment 40

Maximum shear 31

Metals in shear 16

Modulus of rupture 72

Moment diagrams 36

Moment of a force . . 16

INDEX 125

Page

Moment of inertia. 46

of built-up sections 49

of a rectangle 48

unit of 47

Moments, principle of 17

Neutral axis 54

Neutral line 54

Neutral surface 54

Non-elastic deformation 117

Notation 24, 33

Parabola-Euler formulas 98

Posts 85

Principle of moments 17

applied to areas 42

Radius of gyration 86

Rankine's column formula 88

Reactions of supports 16

Rectangle, moment of inertia of 48

Reduction formula 48

Resisting moment 57, 107

value of 58

Resisting shear 73

Restrained beam 19

Rivet, shearing strength of 118

Riveted joints 118

Rods, stiffness of 110

Rule of signs . . .23, 32

Safe load of a beam 71

Safe strength of a beam 71

Second beam formula 73

Section modulus 59

Shafts

stiffness of 110

strength of 105

twist of 116

twisting moment of 105

Shear diagrams 27

Shear stress 2

Shearing strength of rivet 118

Shears, unit for , 24

Signs, rule of 23, 32

Simple beam 19

Simple stress 1

Single shear 119

Steel in compression 14

Steel in tension 12

Stiffness of rods, beams, and shafts 110

Stone, ultimate strength of 15

126 INDEX

Page

Straight-line formulas 94

Strength of beams 52

laws of 71

Strength of columns 85

Strength of joint, computation of 120

Strength of shafts 105

Stress, kinds of 2

Stress-deformation diagram 6

Struts 85

Tables

bending moment 53

coefficients of elasticity Ill

deflection 53

factors of safety 10

maximum shear, values of 53

mild steel columns, data for 95, 99

moduli of rupture 72

moment diagrams 53

moments of inertia 52

properties of standard I-beams 70

radii of gyration 52

section moduli 52

shear diagrams 53

Temperature stresses 113

Tensile strength of riveted plates 120

Tension, materials in 11

Timber

in compression 14

in shear 15

in tension 11

Timber beams, design of 76

Torsional stress 106

Transverse forces 52

Twist of shafts 116

Twisting moment of shaft 105

Ultimate strength 6

Unit-deformation 4

Unit of moment of inertia 47

Unit-stress 3

Units 33

Units for shears 24

Value of resisting moment 58

Wrought-iron

in compression 14

in tension 12

Working strength

Working stress 8

THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW

AN INITIAL FINE OF 25 CENTS

WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $t.OO ON THE SEVENTH DAY
OVERDUE.

«OV 6 1941 M.

LD 21-100m-7,'40 (6936s)

3S294G

UNIVERSITY OF CALIFORNIA LIBRARY

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Full text of "Strength of materials; a comprehensive presentation of scientific methods of locating and determining stresses and calculating the required strength and dimensions of building materials"

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